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6.1.1: Practice Problems- Solution Concentration

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PROBLEM \(\PageIndex{1}\)

Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.

The number of moles always stays the same in a dilution.

The concentration and the volumes change in a dilution.

PROBLEM \(\PageIndex{2}\)

What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?

The two samples contain the same proportion of moles of salt to liters of solution, but have different numbers of actual moles.

PROBLEM \(\PageIndex{3}\)

Determine the molarity for each of the following solutions:

• 0.444 mol of CoCl 2 in 0.654 L of solution
• 98.0 g of phosphoric acid, H 3 PO 4 , in 1.00 L of solution
• 0.2074 g of calcium hydroxide, Ca(OH) 2 , in 40.00 mL of solution
• 10.5 kg of Na 2 SO 4 ·10H 2 O in 18.60 L of solution
• 7.0 × 10 −3 mol of I 2 in 100.0 mL of solution
• 1.8 × 10 4 mg of HCl in 0.075 L of solution

PROBLEM \(\PageIndex{4}\)

Determine the molarity of each of the following solutions:

• 1.457 mol KCl in 1.500 L of solution
• 0.515 g of H 2 SO 4 in 1.00 L of solution
• 20.54 g of Al(NO 3 ) 3 in 1575 mL of solution
• 2.76 kg of CuSO 4 ·5H 2 O in 1.45 L of solution
• 0.005653 mol of Br 2 in 10.00 mL of solution
• 0.000889 g of glycine, C 2 H 5 NO 2 , in 1.05 mL of solution

5.25 × 10 -3 M

6.122 × 10 -2 M

1.13 × 10 -2 M

PROBLEM \(\PageIndex{5}\)

Calculate the number of moles and the mass of the solute in each of the following solutions:

(a) 2.00 L of 18.5 M H 2 SO 4 , concentrated sulfuric acid (b) 100.0 mL of 3.8 × 10 −5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum (c) 5.50 L of 13.3 M H 2 CO, the formaldehyde used to “fix” tissue samples (d) 325 mL of 1.8 × 10 −6 M FeSO 4 , the minimum concentration of iron sulfate detectable by taste in drinking water

37.0 mol H 2 SO 4

3.63 × 10 3 g H 2 SO 4

3.8 × 10 −6 mol NaCN

1.9 × 10 −4 g NaCN

73.2 mol H 2 CO

2.20 kg H 2 CO

5.9 × 10 −7 mol FeSO 4

8.9 × 10 −5 g FeSO 4

PROBLEM \(\PageIndex{6}\)

Calculate the molarity of each of the following solutions:

(a) 0.195 g of cholesterol, C 27 H 46 O, in 0.100 L of serum, the average concentration of cholesterol in human serum (b) 4.25 g of NH 3 in 0.500 L of solution, the concentration of NH 3 in household ammonia (c) 1.49 kg of isopropyl alcohol, C 3 H 7 OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol (d) 0.029 g of I 2 in 0.100 L of solution, the solubility of I 2 in water at 20 °C

5.04 × 10 −3 M

1.1 × 10 −3 M

PROBLEM \(\PageIndex{7}\)

There is about 1.0 g of calcium, as Ca 2+ , in 1.0 L of milk. What is the molarity of Ca 2+ in milk?

PROBLEM \(\PageIndex{8}\)

What volume of a 1.00- M Fe(NO 3 ) 3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M ?

PROBLEM \(\PageIndex{9}\)

If 0.1718 L of a 0.3556- M C 3 H 7 OH solution is diluted to a concentration of 0.1222 M , what is the volume of the resulting solution?

PROBLEM \(\PageIndex{10}\)

What volume of a 0.33- M C 12 H 22 O 11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M ?

PROBLEM \(\PageIndex{11}\)

What is the concentration of the NaCl solution that results when 0.150 L of a 0.556- M solution is allowed to evaporate until the volume is reduced to 0.105 L?

PROBLEM \(\PageIndex{12}\)

What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?

• 1.00 L of a 0.250- M solution of Fe(NO 3 ) 3 is diluted to a final volume of 2.00 L
• 0.5000 L of a 0.1222- M solution of C 3 H 7 OH is diluted to a final volume of 1.250 L
• 2.35 L of a 0.350- M solution of H 3 PO 4 is diluted to a final volume of 4.00 L
• 22.50 mL of a 0.025- M solution of C 12 H 22 O 11 is diluted to 100.0 mL

PROBLEM \(\PageIndex{13}\)

What is the final concentration of the solution produced when 225.5 mL of a 0.09988- M solution of Na 2 CO 3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?

PROBLEM \(\PageIndex{14}\)

A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?

PROBLEM \(\PageIndex{15}\)

An experiment in a general chemistry laboratory calls for a 2.00- M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?

PROBLEM \(\PageIndex{16}\)

What volume of a 0.20- M K 2 SO 4 solution contains 57 g of K 2 SO 4 ?

PROBLEM \(\PageIndex{17}\)

The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium , which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K 2 Cr 2 O 7 ), what is the maximum permissible molarity of that substance?

4.8 × 10 −6 M

Contributors

Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.  Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ).

• Adelaide Clark, Oregon Institute of Technology

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Concentration of solutions: definitions, formulas, solved problems.

Analytical Chemistry , General Chemistry

– In this subject, we will discuss the Concentration of Solutions Concentration of Solutions (Definitions, Formulas, Solved Problems).

Concentration of Solutions

– For history, measurements and their corresponding units were invented at the local level.

– By necessity of primitive communication and local technology, standards were nearly nonexistent, and conversions among the many systems were difficult.

– The result was many hundreds of distinct ways of expressing concentrations of solutions.

– Fortunately for us, the advent of rapid communications technology and the development of efficient travel have forced the globalization of measurement science and, along with it, the definition of global measurement standards.

– No field has enjoyed more benefit in this regard than chemistry in general and analytical chemistry in particular.

– Even so, we use several methods for expressing concentration.

Ways for Expressing Concentration of Solutions

– In this subject, we describe the four fundamental ways of expressing solution concentration:

(1) molar concentration

(2) percent concentration

(3) solution-diluent volume ratio

(4) p-functions.

(1) Molar Concentration

– The molar concentration c x of a solution of a solute species X is the number of moles of that species that is contained in 1 liter of the solution (not 1 L of the solvent).

– In terms of the number of moles of solute, n, and the volume, V, of solution, we write:

– The unit of molar concentration is molar, symbolized by M, which has the dimensions of mol/L, or mol L -1 .

– Molar concentration is also the number of millimoles of solute per milliliter of solution.

Example (1): Calculate the molar concentration of ethanol in an aqueous solution that contains 2.30 g of C 2 H 5 OH (46.07 g/mol) in 3.50 L of solution.

– To calculate molar concentration, we must find both the amount of ethanol and the volume of the solution.

– The volume is given as 3.50 L, so all we need to do is convert the mass of ethanol to the corresponding amount of ethanol in moles:

– To obtain the molar concentration, c C2H5OH , we divide the amount by the volume. Thus:

–  We will see that there are two ways of expressing molar concentration:

• molar analytical concentration
• molar equilibrium concentration.

– The distinction between these two expressions is in whether the solute undergoes a chemical change in the solution process.

(A) Molar Analytical Concentration

– The molar analytical concentration, or for the sake of brevity, just analytical concentration, of a solution gives the total number of moles of a solute in 1 liter of the solution (or the total number of millimoles in 1 mL).

– In other words, the molar analytical concentration specifies a recipe by which the solution can be prepared regardless of what might happen to the solute during the solution process.

– Note that in Example (1), the molar concentration that we calculated is also the molar analytical concentration c C2H5OH = 0.0143 M because the solute ethanol molecules are intact following the solution process.

– In another example, a sulfuric acid solution that has an analytical concentration of c H2SO4 = 1.0 M can be prepared by dissolving 1.0 mole, or 98 g, of H 2 SO 4 in water and diluting the acid to exactly 1.0 L.

– As we shall see, there are important differences between the ethanol and sulfuric acid examples.

(B) Molar Equilibrium Concentration

– The molar equilibrium concentration, or just equilibrium concentration, refers to the molar concentration of a particular species in a solution at equilibrium.

– To specify the molar equilibrium concentration of a species, it is necessary to know how the solute behaves when it is dissolved in a solvent.

– For example, the molar equilibrium concentration of H 2 SO 4 in a solution with a molar analytical concentration c H2SO4 = 1.0 M is actually 0.0 M because the sulfuric acid is completely dissociated into a mixture of H + , HSO 4 – , SO 4 – ions.

– There are essentially no H 2 SO 4 molecules in this solution.

– The equilibrium concentrations of the ions are 1.01, 0.99, and 0.01M, respectively.

– Equilibrium molar concentrations are usually symbolized by placing square brackets around the chemical formula for the species.

– So, for our solution of H 2 SO 4 with an analytical concentration of c H2SO4 = 1.0 M, we write:

Solved Problems on Molar Concentration

Example (2): Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl 3 CCOOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized in water).

– As in Example (1), we calculate the number of moles of Cl 3 CCOOH, which we designate as HA, and divide by the volume of the solution, 10.0 mL, or 0.0100 L. Therefore:

– The molar analytical concentration, c HA , is then:

– In this solution, 73% of the HA dissociates, giving H + and A – :

– The equilibrium concentration of HA is then 27% of c HA . Thus,

– The equilibrium concentration of A – is equal to 73% of the analytical concentration of HA, that is

– Because 1 mole of H + is formed for each mole of A – , we can also write:

Example (3): Describe the preparation of 2.00 L of 0.108 M BaCl 2 from BaCl 2 .2H 2 O (244.3 g/mol).

– To determine the number of grams of solute to be dissolved and diluted to 2.00 L, we note that 1 mole of the dihydrate yields 1 mole of BaCl 2 .

– Therefore, to produce this solution we will need:

– The mass of BaCl 2 .2H 2 O is then:

– Dissolve 52.8 g of BaCl 2 .2H2O in water and dilute to 2.00 L.

Example (4): Describe the preparation of 500 mL of 0.0740 M Cl 2 solution from solid BaCl 2 .2H 2 O (244.3 g/mol).

– Dissolve 4.52 g of BaCl 2 .2H 2 O in water and dilute to 0.500 L or 500 mL.

(2) Percent Concentration

– Chemists frequently express concentrations in terms of percent (parts per hundred).

– Unfortunately, this practice can be a source of ambiguity because the percent composition of a solution can be expressed in several ways.

– Three common methods are:

(1) the denominator in each of these expressions is the mass or volume of solution rather than the mass or volume of the solvent.

(2) the first two expressions do not depend on the units used for weight (mass) as long as the same units are used in the numerator and the denominator.

(3) In the third expression, units must be defined because the numerator and denominator have different units that do not cancel.

(4) Of the three expressions, only weight percent has the advantage of being temperature independent.

– Weight percent is often used to express the concentration of commercial aqueous reagents.

– For example, nitric acid is sold as a 70% (w/w) solution, meaning that the reagent contains 70 g of HNO 3 per 100 g of solution (see Example 8).

– Volume percent is commonly used to specify the concentration of a solution prepared by diluting a pure liquid compound with another liquid.

– For example, a 5% (v/v) aqueous solution of methanol usually describes a solution prepared by diluting 5.0 mL of pure methanol with enough water to give 100 mL.

– Weight or volume percent is often used to indicate the composition of dilute aqueous solutions of solid reagents.

– For example, 5% (w/v) aqueous silver nitrate often refers to a solution prepared by dissolving 5 g of silver nitrate in sufficient water to give 100 mL of solution.

– To avoid uncertainty, always specify explicitly the type of percent composition being discussed.

– If this information is missing, the investigator must decide intuitively which of the several types is to be used.

– The potential error resulting from a wrong choice is considerable.

– For example, commercial 50% (w/w) sodium hydroxide contains 763 g NaOH per liter, which corresponds to 76.3% (w/v) sodium hydroxide.

Parts per Million and Parts per Billion

– For very dilute solutions, parts per million ( ppm ) is a convenient way to express concentration:

– where c ppm is the concentration in parts per million.

– The units of mass in the numerator and denominator must agree so that they cancel.

– For even more dilute solutions, 10 9 ppb rather than 10 6 ppm is used in the previous equation to give the results in parts per billion (ppb).

– The term parts per thousand (ppt) is also used, especially in oceanography.

Example (5): What is the molar concentration of K + in a solution that contains 63.3 ppm of K 3 Fe(CN) 6 (329.3 g/mol)?

– Because the solution is so dilute, it is reasonable to assume that its density is 1.00 g/mL. therefore:

(3) Solution-Diluent Volume Ratios

– The composition of a dilute solution is sometimes specified in terms of the volume of a more concentrated solution and the volume of solvent used in diluting it.

– The volume of the former is separated from that of the latter by a colon.

– Thus, a 1:4 HCl solution contains four volumes of water for each volume of concentrated hydrochloric acid.

– This method of notation is frequently ambiguous in that the concentration of the original solution is not always obvious to the reader.

– Moreover, under some circumstances, 1:4 means diluting one volume with three volumes.

– Because of such uncertainties, you should avoid using solution-diluent ratios.

(4) p-Functions

– Scientists frequently express the concentration of a species in terms of its p-function, or p-value.

– The p-value is the negative logarithm (to the base 10) of the molar concentration of that species. Thus, for the species X:

As shown by the following examples, p-values offer the advantage of allowing concentrations that vary over ten or more orders of magnitude to be expressed in terms of small positive numbers.

Example (6): Calculate the p-value for each ion in a solution that is 2.00 × 10 -3 M in NaCl and 5.4 × 10 -4 M in HCl.

– To obtain pNa, we write:

– The total Cl 2 concentration is given by the sum of the concentrations of the two solutes:

Example (7): Calculate the molar concentration of Ag + in a solution that has a pAg of 6.372.

Density and Specific Gravity of Solutions

– Density and specific gravity are related terms often found in the analytical literature.

– The density of a substance is its mass per unit volume, and its specific gravity is the ratio of its mass to the mass of an equal volume of water at 4°C.

– Density has units of kilograms per liter or grams per milliliter in the metric system. Specific gravity is dimensionless and so is not tied to any particular system of units. For this reason, specific gravity is widely used in describing items of commerce.

– The following Figure shows the Label from a bottle of reagent-grade hydrochloric acid.

– Note that the specific gravity of the acid over the temperature range of 60° to 80°F is specified on the label:

– Since the density of water is approximately 1.00 g/mL and since we use the metric system throughout this text, we use density and specific gravity interchangeably.

– The specific gravities of some concentrated acids and bases are given in the following Table:

Solved Problems on Concentration 

Example (8): Calculate the molar concentration of HNO 3 (63.0 g/mol) in a solution that has a specific gravity of 1.42 and is 70.5% HNO 3 (w/w).

Let us first calculate the mass of acid per liter of concentrated solution:

Example (9): Describe the preparation of 100 mL of 6.0 M HCl from a concentrated solution that has a specific gravity of 1.18 and is 37% (w/w) HCl (36.5 g/mol).

– Proceeding as in Example (8), we first calculate the molar concentration of the concentrated reagent.

– We then calculate the number of moles of acid that we need for the diluted solution.

– Finally, we divide the second figure by the first to obtain the volume of concentrated acid required.

– Thus, to obtain the concentration of the reagent, we write:

– The number of moles HCl required is given by:

– Finally, to obtain the volume of concentrated reagent, we write:

– Therefore, dilute 50 mL of the concentrated reagent to 600 mL.

– The solution to Example (9) is based on the following useful relationship, which we will be using countless times:

– where the two terms on the left are the volume and molar concentration of a concentrated solution that is being used to prepare a diluted solution having the volume and concentration given by the corresponding terms on the right.

– This equation is based on the fact that the number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.

– Note that the volumes can be in milliliters or liters as long as the same units are used for both solutions.

Reference: Fundamentals of analytical chemistry / Douglas A. Skoog, Donald M. West, F. James Holler, Stanley R. Crouch. (ninth edition), 2014. USA

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Concentration of Solutions

This is a series of lectures in videos covering Chemistry topics taught in High Schools. These lessons look at calculating the concentration of a solution.

Related Pages IGCSE Chemistry High School Chemistry More Lessons for Chemistry

Concentrations Calculations (% weight/volume) A tutorial on calculating the concentration of a solution in % weight-volume.

Example: You dissolve 2.5 moles of strantium acetate into 1.0 L of water. Determine the concentration of the solution in %W/C.

Concentration of Solutions: Volume/Volume % (v/v) A volume/volume percent (v/v) gives the volume of solute divided by the volume of the solution (expressed as a percent). The following video looks at calculating concentration of solutions. We will look at a sample problem dealing with volume/volume percent (v/v)%

Example: Rubbing alcohol is commonly used as an antiseptic for small cuts. It is sold as 70% (v/v) solution of isopropyl alcohol in water. What volume of isopropyl alcohol is used to make 500 mL of rubbing alcohol?

Concentration of Solutions: mass/volume % (m/v)% The following video looks at calculating concentration of solutions. We will look at a sample problem dealing with mass/volume percent (m/v)%.

Example: Many people use a solution of sodium phosphate (Na 3 PO 4 - commonly called TSP), to clean walls before putting up wallpaper. The recommended concentration is 1.7%(m/v). What mass of TSP is needed to make 2.0 L of solution?

Concentration of Solutions: Mass/Mass % (m/m)% A mass/mass percent gives the mass of a solute divided by the mass of solution (expressed as a percent) The following video looks at calculating concentration of solutions. We will look at a sample problem dealing with mass/mass percent (m/m)%

Example: CaCl 2 is used to melt ice on roads. To determine how much CaCl 2 has been used, you take a sample of slush to analyze. The sample had a mass of 23.47g. When the solution was evaporated, the residue had a mass of 4.58g. What was the mass/mass percent of CaCl 2 in the slush? How many grams of CaCl 2 were present in 100g of solution?

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8.010 m means 8.010 mol / 1 kg of solvent (8.010 mol) (98.0768 g/mol) = 785.6 g of solute 785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution. 1785.6 g / 1.354 g/mL = 1318.76 mL 8.01 moles / 1.31876 L = 6.0739 M 6.074 M (to four sig figs)

1 L of solution = 1000 mL = 1000 cm 3 1.329 g/cm 3 times 1000 cm 3 = 1329 g (the mass of the entire solution) 1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution) 571.4 g / 98.0768 g/mol = 5.826 mol of H 2 SO 4 5.826 mol / 0.7576 kg = 7.690 m

mass of acetone: (3.30 mL) (0.789 g/mL) = 2.6037 g moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol mass of solution: (75.0 mL) (0.993 g/mL) = 74.475 g mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g moles of water: 71.8713 g / 18.015 g/mol = 3.9896 mol

0.04483 mol / 0.0750 L = 0.598 M

0.04483 mol / 0.0718713 kg = 0.624 m

0.04483 mol / (0.04483 mol + 3.9896 mol) = 0.0111

(15.00 mol/L) (1.000 L) = 15.00 mole of HCl 15.00 mol times 36.4609 g/mol = 546.9135 g of HCl

(1000. mL) (1.0745 g/cm 3 ) = 1074.5 g of solution 1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg

15.00 mol / 0.5275865 kg = 28.43 m (to four sig figs)

0.449 mol/kg = x / 2.92 kg x = 1.31108 mol of KBr

(1.31108 mol) (119.0023 g/mol) = 156 g KBr

156 g KBr + 2920 g water = 3076 g total If you wanted to be real technical about it, then use three sig figs to obtain 3080 g.

(0.391 mol) ( 86.1766 g/mol) = 33.6950 g 33.6950 g + 1000 g = 1033.6950 g In other words, every 1033.6950 g of 0.391 m solution delivers 33.6950 g of hexane 33.6950 is to 1033.6950 as 247 is to x x = 7577.46446 g to three sig figs, 7.58 kg of solution

1.42 m means 1.42 mole of C 6 H 6 in 1 kg of tetrahydrofuran (1.42 mol) ( 78.1134 g/mol) = 110.921 g 110.921 g + 1000 g = 1110.921 g 110.921 is to 1110.921 as x is to 1630 x = 162.75 g To check, do this: 162.75 g / 78.1134 g/mol = 2.08351 mol 1630 g − 162.75 g = 1467.25 g 2.08351 mol / 1.46725 kg = 1.42 m

A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900. Let us assume a solution is present made up of 0.100 mole of NaCl and 0.900 mole of water. mass of water present ---> (0.900 mol) (18.015 g/mol) = 16.2135 g molality of solution ---> 0.100 mol / 0.0162135 kg = 6.1677 m to three sig figs, 6.17 m

mass solvent ---> 7550 g − 929 g = 6621 g = 6.621 kg moles solute ---> 929 g/ 84.93 g/mol = 10.9384 mol molality = 10.9384 mol / 6.621 kg = 1.65 m

(1000 mL) (1.230 g/mL) = 1230 g

(3.75 mol) (98.0768 g/mol) = 367.788 g

1230 − 367.788 = 862.212 g

3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs)

1.00 L of this solution contains 4.20 mole of NaCl. (1.00 L) (1050 g/L) = 1050 g of solution.

(4.20 mol) (58.443 g/mol) = 245.4606 g 1050 g - 245.4606 g = 804.5394 g

4.20 mol / 0.8045394 kg = 5.22 m (to three sig figs)

3.58 mole of RbCl in 1000 g of water.

(3.58 mol) (120.921 g/mol) = 432.89718 g 1000 g + 432.89718 g = 1432.89718 g

1432.89718 g / 1.12 g/mL = 1279.37 mL

3.58 mol / 1.27937 L = 2.80 M

molality = moles of naphthalene / kilograms of benzene (16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m

(1.34 mL) (1.59 g/mL) = 2.1306 g 2.1306 g / 153.823 g/mol = 0.013851 mol

(65.0 mL) (1.33 g/mL) = 86.45 g = 0.08645 kg

0.013851 mol / 0.08645 kg = 0.160 m (to three sig figs)

MV = mass / molar mass (x) (0.4500 L) = 0.825 g / 141.9579 g/mol x = 0.0129 M

0.825 g / 141.9579 g/mol = 0.00581158 mol 0.00581158 mol / 0.4500 kg = 0.0129 m

Na 2 HPO 4 ---> 0.825 g / 141.9579 g/mol = 0.00581158 mol H 2 O ---> 450.0 g / 18.015 g/mol = 24.97918401 mol mole fraction of the water ---> 24.97918401 mol / (24.97918401 mol + 0.00581158 mol) = 0.9998

water ---> (450 g / 450.825 g) (100) = 99.8%

ppm means the number of grams of solute per 1,000,000 grams of solution 0.825 is to 450.825 as x is to 1,000,000 x = 1830 ppm

    mol solute m  =  –––––––––     kg solvent

    mass solute n  =  –––––––––––––––––     molar mass of solute

I'm going to use the kg/mol amount and the reason will show up in a moment.     mass solute n  =  ––––––––––––     0.0580794 kg/mol

mass of solution = mass of solvent + mass of solute = 450.0 g = 0.4500 kg I'll ignore those two trailing zeros for the moment. We write this: mass of solvent = 0.45 − mass of solute Here's the reason why I must use the kg/mol unit: In the subtraction just above, the 0.45 is in kilograms. That means the mass of solute must also be in kilograms. You can't subtract two numbers using different units. Also, the bottom unit must be in kilograms because the 0.75 molal value is determined with kg in the denominator. Using grams in the denominator is not done with molality.

  x –––––– 0.0580794 0.75 = ––––––––––––   0.45 − x     x 0.3375 − 0.75x  =  –––––––––     0.0580794 x = 0.019602 − 0.04356x 1.04356x = 0.019602 x = 0.01878 kg = 18.78 g (to 4 sig figs)

0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of water mass of acetone ---> 58.0794 g/mol times 0.7500 mol = 43.56 g mass of solution ---> 1000 g + 43.56 g = 1043.56 g 43.56 is to 1043.56 as x is to 450 x = 18.78 g

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How to Calculate the Concentration of a Solution

Last Updated: March 20, 2024 Fact Checked

This article was co-authored by Chris Hasegawa, PhD and by wikiHow staff writer, Hunter Rising . Dr. Chris Hasegawa was a Science Professor and the Dean at California State University Monterey Bay. Dr. Hasegawa specializes in teaching complex scientific concepts to students. He holds a BS in Biochemistry, a Master’s in Education, and his teaching credential from The University of California, Davis. He earned his PhD in Curriculum and Instruction from The University of Oregon. Before becoming a professor, Dr. Hasegawa conducted biochemical research in Neuropharmacology at the National Institute of Health. He also taught physical and life sciences and served as a teacher and administrator at public schools in California, Oregon, and Arizona. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 2,132,540 times.

In chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. The standard formula is C = m/V, where C is the concentration, m is the mass of the solute dissolved, and V is the total volume of the solution. If you have a small concentration, find the answer in parts per million (ppm) to make it easier to follow. In a lab setting, you may be asked to find the molarity , or molar concentration, of the solution instead.

Using the Mass per Volume Equation

• If the solute you’re using is a liquid, then you can also calculate the mass using the density formula, where density D = m/V, where m is the mass of the liquid and V is the volume. To find the mass, multiply the density of the liquid by the volume.

Tip: If you need to use a scale, subtract the mass of the container you’re using to hold the solute or else your calculations will be off.

• If you aren’t measuring the volume yourself, you may need to convert the mass of the solute into volume using the density formula.
• For example, if you’re finding the concentration of 3.45 grams of salt in 2 liters of water, you would find the volume of salt using the density formula. Look up the density of salt either in a textbook or online and solve the formula for m. In this case, the density of salt is 2.16 g/mL. The formula would read 2.16 g/mL = (3.45 g)/V. Multiply each side by V to get V(2.16 g/mL) = 3.45 g. Then divide the each side by 2.16 to find the volume, or V = (3.45 g)/(2.16 g/mL) = 1.60 mL.
• Add the volume of the solute to the volume of your solvent, ma. So in this example, 2 L + 1.6 mL = 2,000 mL + 1.6 mL = 2,001.6 mL. You can either leave the measurement in milliliters or convert it back to liters to get 2.002 L.

• In our example for the concentration of 3.45 grams of salt in 2 liters of water, your equation would be C = (3.45 g)/(2.002 L) = 1.723 g/L.
• Certain problems may ask for your concentration in specific units. Be sure to convert the units before putting them in your final formula.

Finding Concentration in Percentage or Parts per Million

• If your solute is a liquid, you may need to calculate the mass using the formula D = m/V, where D is the liquid’s density, m is the mass, and V is the volume. Look up the density of the liquid in a textbook or online and then solve the equation for the mass.

• For example, if you want to find the concentration of 10 g of cocoa powder mixed with 1.2 L of water, you would find the mass of the water using the density formula. The density of water is 1,000 g/L, so your equation would read 1,000 g/L = m/(1.2 L). Multiply each side by 1.2 L to solve the mass in grams, so m = (1.2 L)(1,000 g/L) = 1,200 g. Add the mass of the cocoa powder to get 1,210 g.

• In our example, C = (10 g)/(1,210 g) = 0.00826.

• In this example, the percent concentration is (0.00826)(100) = 0.826%.

• In our example, the ppm = (0.00826)(1,000,000) = 8,260 ppm.

Tip: Parts per million is usually used for very small concentrations since it’s easier to write and understand than a percentage.

Calculating Molarity

• For example, if your solute is potassium hydroxide (KOH), find the atomic masses for potassium, oxygen, and hydrogen and add them together. In this case molar mass = 39 +16 + 1 = 56 g/mol.
• Molarity is used mainly in chemistry when you know the chemical makeup of the solute you’re using.

• For example, if you want to find the number of moles in 25 g of potassium hydroxide (KOH), then the equation is mol = (25 g)/(56 g/mol) = 0.45 mol
• Convert the mass of your solute to grams if it isn’t already listed in grams.
• Moles are used to represent the number of atoms in the solution.

• In this example, if you’re using 400 mL of water, then divide it by 1,000 to convert it to liters, which is 0.4 L.
• If your solvent is already listed in liters, then you can skip this step.

Tip: You don’t need to include the volume of the solute since it doesn’t usually affect the volume that much. If there is a visible change in volume when you mix the solute with the solvent, then use the total volume instead.

• In this example, M = (0.45 mol)/(0.4 L) = 1.125 M.

Calculator, Practice Problems, and Answers

Community Q&A

• If you are in a lab and don’t know how much of a solute was added, you can perform a titration test using other reactive chemicals. You do need to learn how to balance chemical equations with stoichiometry . Thanks Helpful 0 Not Helpful 0

You Might Also Like

• ↑ https://www.physiologyweb.com/calculators/mass_per_volume_solution_concentration_calculator.html
• ↑ https://www.omnicalculator.com/conversion/ppm
• ↑ https://sciencing.com/calculate-concentration-ppm-6935286.html
• ↑ https://chem.libretexts.org/Courses/Los_Angeles_Trade_Technical_College/Chem_51/15%3A_Solutions/15.03%3A_Solution_Concentration_-_Molality_Mass_Percent_ppm_and_ppb
• ↑ https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/04%3A_Reactions_in_Aqueous_Solution/4.05%3A_Concentration_of_Solutions
• ↑ Chris Hasegawa, PhD. Retired Science Professor & Dean. Expert Interview. 29 July 2021.
• ↑ https://www.khanacademy.org/science/chemistry/states-of-matter-and-intermolecular-forces/mixtures-and-solutions/a/molarity
• ↑ https://www.inchcalculator.com/convert/milliliter-to-liter/

About This Article

To calculate the concentration of a solution, start by converting the solute, or the substance being dissolved, into grams. If you're converting from milliliters, you may need to look up the solute's density and then multiply that by the volume to convert to grams. Next, convert the solvent to liters. Finally, divide the solvent by the solute to find the concentration of the solution. To learn how to calculate the concentration of a solution as a percentage or parts per million, scroll down! Did this summary help you? Yes No

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Chemistry: Molarity and Solutions

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Class 9 Chemistry (India)

Course: class 9 chemistry (india)   >   unit 1.

• Types of mixtures

Concentration of a solution

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

• Chemistry Articles
• Concentration Of Solution And The Calculation

Expression of Concentration of Solutions

We always discuss a solution being diluted or concentrated; this is a qualitative way of expressing the concentration of the solution. A dilute solution means the quantity of solute is relatively very small, and a concentrated solution implies that the solution has a large amount of solute. But these are relative terms and do not give us the quantitative concentration of the solution.

Table of Contents

Concentration.

• Recommended Video
• Concentration in parts per million
• Mass percentage
• Volume percentage
• Mass by Volume percentage
• Mole Fraction
• Solutions of Solid in Liquid

Solubility of Gases

Frequently asked questions – faqs.

So, to quantitatively describe the concentrations of various solutions around us, we commonly express levels in the following way:

It is the amount of solute present in one litre of solution. It is denoted by C or S.

Recommended Videos

Introduction to solutions – concentration terms.

Is Matter Around us Pure?

Effect of Change in Concentration

1. Concentration in Parts Per Million (ppm)

The parts of a component per million parts (10 6 ) of the solution.

2. Mass Percentage (w/w):

When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w). Suppose we have a solution containing component A as the solute and B as the solvent, then its mass percentage is expressed as:

Mass % of A = \(\begin{array}{l} \frac {Mass \space of \space component \space A ~ in ~the ~ solution}{Total ~ mass ~of~ the~ solution } × 100 \end{array} \)

3. Volume Percentage (V/V):

Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage and is given as:

volume   % of A = \(\begin{array}{l} \frac {Volume ~of~ component~ A~ in~ the ~solution}{Total ~ volume ~ of ~ the ~ solution } × 100 \end{array} \)

For example, if a solution of NaCl in water is said to be 10 % by volume that means a 100 ml solution will contain 10 ml NaCl.

4. Mass by Volume Percentage (w/V):

This unit is majorly used in the pharmaceutical industry. It is defined as the mass of a solute dissolved per 100mL of the solution.

% w/V = (Mass of component A in the solution/ Total Volume of the Solution)x 100

5. Molarity (M):

One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution. Suppose a solution of ethanol is marked 0.25 M, this means that in one litre of the given solution 0.25 moles of ethanol is dissolved.

Molarity (M) = Moles of Solute/Volume of Solution in litres

6. Molality (m):

Molality represents the concentration regarding moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent .  The molality formula is as given-

\(\begin{array}{l} Molality (m) = \frac {Moles~ of ~solute}{Mass~ of~ solvent~ in~ kg}\end{array} \)

7. Normality

It is the number of gram equivalents of solute present in one litre of the solution and it is denoted by N.

The relation between normality and molarity.

• N x Eq.Wt = Molarity x Molar mass
• N = Molarity x Valency
• N = Molarity x Number of H + or OH –  ion.

8. Formality

It is the number of gram formula units present in one litre of solution. It is denoted by F.

It is applicable in the case of ionic solids like NaCl.

9. Mole Fraction:

If the solution has a solvent and the solute, a mole fraction gives a concentration as the ratio of moles of one component to the total moles present in the solution. It is denoted by x. Suppose we have a solution containing A as a solute and B as the solvent. Let n A and n B be the number of moles of A and B present in the solution respectively. So, mole fractions of A and B are given as:

\(\begin{array}{l}~~~~~~~~~~~~~~~\end{array} \) \(\begin{array}{l} x_A = \frac {n_A}{n_A + n_B}\end{array} \) \(\begin{array}{l}~~~~~~~~~~~~~~~\end{array} \) \(\begin{array}{l} x_B = \frac {n_B}{n_A + n_B} \end{array} \)

The above-mentioned methods are commonly used ways of expressing the concentration of solutions. All the methods describe the same thing that is, the concentration of a solution, each of them has its own advantages and disadvantages. Molarity depends on temperature while mole fraction and molality are independent of temperature. All these methods are used on the basis of the requirement of expressing the concentrations.

Solutions of Solids in Liquids

• A saturated solution is a solution that remains in contact with an excess of solute.
• The amount of solute dissolved per 100g of solvent in a saturated solution at a specific temperature represents the solubility of the solute.
• For exothermic substances such as KOH, CaO, Ca(OH) 2 , M 2 CO 3 , M 2 SO 4 , etc, solubility is inversely proportional to temperature.
• For endothermic substances such as NaCl, KNO 3 , NaNO 3 , glucose, etc., solubility is directly proportional to temperature.

The solubility of gases is mostly expressed in terms of the absorption coefficient,k that is the volume of the gas dissolved by unit volume of solvent at 1 atm pressure and a specific temperature.

The solubility of a gas in a liquid depends upon

• Temperature solubility is inversely proportional to temperature as the dissolution of gas is exothermic in most cases.
• Nature of gas – Gases having a higher value of van der Waals force of attraction that is gases that are more easily liquefied are more soluble. For example, SO 2 and CO 2 are more soluble in water than O 2 , N 2 , and H 2 .
• Nature of solvent – Gases which can ionise in an aqueous solution are more stable in water as compared to the other solvents.

How do you change the concentration of a solution?

Sometimes, by modifying the quantity of solvent, a worker would need to modify the concentration of a solution. Dilution is the addition of a solvent that reduces the solute concentration of the solution. Concentration is solvent elimination, which increases the solute concentration in the solution.

What is a high concentration?

A concentration of persons means that in one place there are more of them. A high concentration of a material in a solution means there’s a lot of it compared to the volume: because of the high concentration of salt, the Great Salt Lake has very little fish.

Is dilute acid dangerous?

In general, a mild condensed acid is more harmful than a solid diluted acid. Although concentrated acetic acid is much less reactive, you do not want to touch the skin or mucous membranes because it is corrosive.

What is the concentration of a solution?

A solution concentration is a measure of the quantity of solute that has been dissolved in a given quantity of solvent or solution. One that contains a relatively high volume of dissolved solute is a concentrated solution. That that contains a relatively minimal volume of dissolved solute is a dilute solution.

How do you prepare a solution of known concentration?

Solutions of known concentration can be prepared either by dissolving the known mass of the solvent solution and diluting it to the desired final volume or by diluting it to the desired final volume by diluting the acceptable volume of the more concentrated solution (the stock solution).

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Chapter 1: Matter in our Surroundings

• Matter is Made of Tiny Particles
• Why Solids, Liquids and Gases Have Different Properties
• Classification of Matter
• Brownian Movement
• States of Matter: Solid, Liquid, Gas and Plasma
• Evaporation
• Effects of Relative Humidity and Wind Speed
• How Does Evaporation Cause Cooling?
• Effect of Change of Temperature
• Melting Point
• What is Vaporization?
• Condensation
• Effects of Change of Pressure
• Difference between Rigidity and Fluidity of Matter
• Prove That Liquids have No fixed Shape but have a Fixed Volume
• Diffusion in Solids, Liquids, and Gases
• What is the Unit of Temperature?
• What is the Relationship Between Celsius and Kelvin Scale of Temperature?
• Liquification of Gases
• How to demonstrate the Presence of Water Vapour in Air?
• What is Plasma and Bose-Einstein Condensate?

Chapter 2: Is Matter Around Us Pure?

• Solution: Properties of Solution
• Saturated and Unsaturated Solutions

Concentration of a Solution

• Suspensions
• How will you distinguish a Colloid from a Solution?
• Classification of Colloids
• Tyndall Effect
• Separation of Mixtures
• How to separate a Mixture of Two Solids?
• Separation by a suitable solvent
• Separation of Mixtures using Sublimation and Magnets
• How to Separate a Mixture of a Solid and a Liquid?
• Filtration: Definition, Process, Diagram and Examples
• Water Purification
• Centrifugation
• How to Separate Cream from milk?
• Difference Between Homogeneous and Heterogeneous Mixture
• Difference Between Compound and Mixture
• Factors affecting Solubility
• Separation by Evaporation
• Crystallization
• Chromatography
• Distillation
• Separation of Mixtures of Two or More Liquids
• Fractional Distillation
• Pure and Impure Substances
• What is an Element?
• Metals, Non-Metals and Metalloids
• Properties of Metals and Non-Metals

Chapter 3: Atoms and Molecules

• Laws of Chemical Combination
• Law of Conservation of Mass
• Verification of the Law of Conservation of Mass in a Chemical Reaction
• Law of Constant Proportions
• What is Atom?
• Atomic Mass
• How Do Atoms Exist?
• Cations vs Anions
• What are Ionic Compounds?
• What are Monovalent Ions?
• What are Divalent Ions?
• Trivalent Ions - Cations and Anions
• Polyatomic Ions
• Formulas of Ionic Compounds
• Chemical Formula
• Chemical Formula of Common Compounds
• Molecular Mass
• Mole Concept
• Problems Based on Mole Concepts
• Dalton's Atomic Theory
• Drawbacks of Dalton's Atomic Theory
• Significance of the Symbol of Elements
• Difference Between Molecules and Compounds
• How to Calculate Valency of Radicals?
• What is the Significance of the Formula of a Substance?
• Gram Atomic and Gram Molecular Mass

Chapter 4: Structure of the Atom

• Charged Particles in Matter
• Thomson's Atomic Model
• Rutherford Atomic Model
• Drawbacks of Rutherford's Atomic Model
• Bohr's Model of an Atom
• Valence Electrons
• Mass Number
• Relation Between Mass Number and Atomic Number
• Why do all the Isotopes of an Element have similar Chemical Properties?
• Why Isotopes have different Physical Properties?
• What is Fractional Atomic Mass?
• Radioactive Isotopes
• Discovery of Electrons
• What is a Proton?
• Rutherford's Alpha Scattering Experiment
• Atomic Nucleus
• How did Neil Bohr explained the Stability of Atom?
• Electron Configuration
• Potassium and Calcium - Atomic Structure, Chemical Properties, Uses
• What is meant by Chemical Combination?
• Difference between Electrovalency and Covalency

Concentration of Solution is a measure of the amount of solute that has been dissolved in the given amount of solvent. In simple words, it means determining how much of one substance is mixed with another substance. As Concentration is a frequently used term in chemistry and other relevant fields, although it is most commonly used in the context of solutions, where it refers to the quantity of solute dissolved in a solvent. Concentration can be expressed in both qualitative or quantitative (numerically) terms.

Concentration of a Solution Definition

Concentration of a solution is defined as the amount of solute dissolved in the solution. It is given by the ratio of the amount of solute to the amount of solution or solvent sometimes. However, we can express it in percentages, Parts per Million, and several other ways. The concentration of a solution can be expressed both qualitatively and quantitatively which we will see in the below topics. Before learning more about concentration let’s understand some of the general types of solution

Solution of Solid and Liquid

Solutions of solids and liquids involve the dissolution of a solid solute in a liquid solvent or a liquid solute in a liquid solvent. Examples include:

a. Solid in Liquid: Dissolving table salt (NaCl) in water.

b. Liquid in Liquid: Mixing ethanol and water to form a homogeneous solution.

Solution of Gas

Solutions of gases involve the dissolution of a gas in a liquid solvent or another gas. Examples include:

a. Gas in Liquid: Dissolving carbon dioxide (CO 2 ) in water to form carbonated water.

b. Gas in Gas: Mixing different gases, such as oxygen and nitrogen, in the atmosphere to form a homogeneous mixture.

Qualitative Expressions of Concentration

To qualitatively express concentration, a solution can be classified as a dilute solution or a concentrated solution, which are explained as follows:

Dilute Solution

Dilute Solution contains a smaller proportion of solute than the proportion of solvent. For example, if 2 grams of salicylic acid is dissolved in 100 ml of water and in another container, 8 grams of salicylic acid is dissolved in the same amount of water then a 2-gram solution of salicylic acid is a dilute solution compared to 8 grams solution of salicylic acid.

Concentrated Solution

Concentrated Solution contains a much greater proportion of solute than the proportion of solvent. For example, if 2 grams of salicylic acid is dissolved in 100 ml of water and in another container, 8 grams of salicylic acid is dissolved in the same amount of water then 8 grams solution of salicylic acid is a concentrated solution compared to 2 grams solution of salicylic acid.

Figure 1. Dilute (left) and Concentrated (right) solutions

Semi-Qualitative Expressions of Concentration

To semi-qualitatively express concentration, a solution can be classified as a saturated solution or an unsaturated solution , which are explained as follows:

Saturated Solution  

A saturated solution is one in which the greatest quantity of solute is dissolved in a solvent at a given temperature. When a solution reaches saturation, it can no longer dissolve any more solute at that temperature. Undissolved chemicals settle to the bottom. The saturation point is determined as the point at which no more solute can be dissolved in the solvent.

Unsaturated Solution

An unsaturated solution is one that contains less solute than the maximum possible solute it can dissolve before the solution reaches the saturation level. When more solute is dissolved in this solution, there are no residual substances at the bottom, indicating that all of the solutes have been dissolved in the solvent. An unsaturated solution is a chemical solution in which the solute concentration is less than the corresponding equilibrium solubility.

Figure 2. Unsaturated (left) and Saturated (right) solutions

Solubility is defined as the greatest amount of solute that may dissolve in a certain quantity of solvent at a given temperature.

A solution is a liquid that is a homogeneous combination of one or more solutes and a solvent. A typical example of a solution is, sugar cubes added to a cup of tea or coffee. Here, solubility is the characteristic that allows sugar molecules to dissolve.

As a result, the term solubility may be defined as a substance’s (solute’s) ability to dissolve in a particular solvent.

Quantitative Expressions of Concentration

Qualitative expressions of concentration are relative terms, which do not provide the exact concentration of the solution. To characterize the concentrations of various solutions around us in an accurate and precise manner, we require quantitative expressions of concentration.

Generally, concentration is represented in both ways: Concentration = Quantity of Solute / Quantity of Solution or Concentration = Quantity of Solute / Quantity of Solvent

To quantitatively express concentration, we use the following terms:

• Mass Percentage
• Volume Percentage
• Mass by Volume Percentage
• Parts per Million and Parts per Billion

Mole Fraction 

Mass percentage (w/w%).

Mass percentage which is also called weight by weight concentration of solute and is defined as the amount of solute (in grams) present in 100 gm of the solution.

Mass Percentage = (Mass of Solute / Mass of Solution) × 100

Mass percentage has no unit as it is the ratio of the mass of solute and solution.

Volume Percentage (v/v%)

Volume Percentage which is also called volume by volume concentration of solute and is defined as the amount of solute (in ml) present in 100 ml of the solution.

Volume Percentage  = (Volume of Solute / Volume of Solution) × 100

Volume percentage has no unit as it is the ratio of the volume of solute and solution.

Mass by Volume Percentage(w/v %)

It is defined as the amount of solute (in grams) present in the 100ml of the solution.

Mass by Volume Percentage = (Mass of Solute(in gm) / Volume of Solution(in ml) × 100

Unit of mass by volume percentage is gram per milliliter as it is the ratio of the mass of the solute and volume of the solution.

Parts per Million (PPM)

Parts Per Million or PPM is used to measure the very small amount of solute dissolved in the Solvent. Drinking water, air, soils, etc. are the solvents that have very fewer amounts of solutes in them, which can’t be measured in percentage as a percentage only calculates the concentration out of 100. If we represent concentrations of these solvents in percentage it looks like 0.00002 %, which is not an effective way. That’s why parts per million were introduced to make a representation of these concentrations.

PPM of solute = Mass of solute (in milligrams)/Mass of solution(In Kg)

Parts per Billion

Like, Parts per million, Parts Per Billion are also used to represent solute available in trace quantities. Parts Per Billion represents the amount of solute in 1 billion parts of the solution. 

PPB of solute = Mass of solute (in micrograms)/Mass of solution(In Kg)

Molarity of a given solution is defined as the number of moles present in the 1 liter of solution. For example, if 2 moles of NaCl are dissolved in 1 liter of water, the molarity of the resulting solution would be 2M, and the Formula for Molarity is given as follows:

Molarity (M) = Moles of solute /Volume of solution(in Liter)

Molality of a given solution is defined as the number of moles present in 1 kg of solution. For example, if 3 moles of KOH are dissolved in 3 Liters of water (density of water 1 kg/L), the molality of the resulting solution would be 1 m, as there is 1 mole of KOH present in each Kg of water. The formula for molality is given as follows:

Molality (m) = Moles of solute / Mass of solvent(in Kg)

Mole fraction i.e., X is defined as the ratio of the number of moles of one component to the total number of moles present in the solution. It is a dimensionless quantity. The mole fraction of solute A in a solution containing n components such as A, B, C, . . ., N can be calculated using the following formula:

Mole fraction of A (X A ) = Moles of A / (Moles of A + Moles of B + . . . + Moles of N)

The mole fraction of other solvents (B, C, D, . . .N) in a solution can be calculated using a similar formula.

Normality is a measure of concentration equivalent to the number of equivalents per liter of solution. It is often used for reactions that involve acid-base neutralization, precipitation reactions, or redox reactions, and it takes into account the stoichiometry of the reaction.

Normality (N) = Equivalents of solute / Volume of solution in liters

For example, for an acid-base reaction, normality can be calculated using:

Normality = Molarity × Basicity or Acidity of the compound

Where basicity or acidity is the number of hydrogen ions (H⁺) or hydroxide ions (OH⁻) that can be released per molecule of the compound.

Formality is similar to molarity in that it measures the number of moles of solute per liter of solution. However, formality is used when the solute undergoes a reaction or dissociation in solution. Formality measures the concentration based on the initial composition of the solution, not the final dissolved state.

Formality (F) = Moles of solute / Volume of solution in liters

Temperature Dependence of Quantitative Expressions of Concentration

The following table shows the temperature dependence of the Quantitative Expressions of Concentration.

Mole Concept Heterogeneous and Homogeneous Mixtures Ideal and Non-Ideal Solutions

Sample Problems on Concentration of Solution

Problem 1: 15 g of common salt is dissolved in 400 g of water. Calculate the concentration of the solution by expressing it in Mass by Mass percentage (w/w%).

Given that, Mass of solute (common salt) = 15 g    …(1) Mass of Solvent (water) = 400 g    …(2) It is known that,  Mass of Solution = Mass of Solute + Mass of Solvent    …(3) So,  Substituting (1) and (2) in (3), we obtain the following, Mass of Solution = 15 g + 400 g = 415 g    …(4) From Figure 4, we know Mass by Mass Percentage = ( Mass of Solute / Mass of Solution ) × 100    …(5) Substituting (1) and (4) in (5), we obtain the following, Mass by Mass Percentage = ( 15 g / 415 g ) × 100 = ( 0.0361 ) × 100 = 3.61 Answer is: ( w / w % ) = 3.61

Problem 2: 15 g of common salt is dissolved in a solution of 300 mL, calculate the Mass by Volume percentage (w/v%).

Given that, Mass of solute (common salt) = 15 g     . . . (1) Mass of Solution (salt solution) = 300 mL    . . . (2) From Figure 5, we know Mass by Volume Percentage = ( Mass of Solute / Volume of Solution ) × 100   . . . (3) Substituting (1) and (2) in (3), we obtain the following, Mass by Volume Percentage = ( 15 g / 300 mL ) × 100 = ( 0.05 ) × 100 = 5 g/mL Answer is: ( w / v % ) = 5 g/mL

Problem 3: Richard dissolved 70 g of sugar in 750 mL of sugar solution. Calculate the Mass by Volume percentage (w/v%).

Given that, Mass of solute (common salt) = 70 g    . . . (1) Mass of Solution (salt solution) = 750 mL     . . . (2) From Figure 5, we know Mass by Volume Percentage = ( Mass of Solute / Volume of Solution ) × 100     . . . (3) Substituting (1) and (2) in (3), we obtain the following, Mass by Volume Percentage = ( 70 g / 750 mL ) * 100 = ( 0.933 ) × 100 = 93.3 g/mL Answer is: ( w / v % ) = 93.3 g/mL

Problem 4: What is the molarity of a solution containing 0.5 moles of NaCl dissolved in 500 mL of water?

Given, Moles of NaCl = 0.5, Volume of Solution = 500 mL = 0.5 Liter As, Molarity (M) = moles of solute / liters of solution ⇒ M = 0.5 moles / 0.5 liters = 1 M So the molarity of the solution is 1 M.

Problem 5: What is the molality of a solution containing 20 g of glucose dissolved in 500 g of water?

Given: Mass of Glucose = 20 g, Mass of water = 500 g = 0.5 kg,  Molar mass of glucose (C 6 H 12 O 6 ) = 180 g/mol Number of Moles = Mass/Molar Mass ⇒ Moles of Glucose = 20 / 180 = 1/9 ≈ 0.111 moles of glucose As, Molality (m) = moles of solute / kilograms of solvent ⇒ m = 0.111 / 0.5 = 0.222 mol/kg So the molality of the solution is 0.222 mol/kg.

Problem 6:How many moles of HCl are present in 250 mL of a 0.2 M HCl solution?

Given: Molarity of solution = 0.2 M, Volume of solution = 250 mL = 0.25 liters Molarity (M) = moles of solute / liters of solution ⇒ Moles of solute = Molarity x liters of solution ⇒ Moles of HCl = 0.2 M x 0.25 L = 0.05 moles So there are 0.05 moles of HCl present in 250 mL of the solution.

Problem 7:What is the ppm of lead in a sample that contains 20 mg of lead in 10 L of water?

Given : mass of solute(in mg) = 20 mg and Volution of solvent = 10 L Mass of solution = Mass of water = 10 L x 1 Kg/L = 10 Kg (density of water is 1Kg/L or 1g/mL) ppm (parts per million) = Mass of solute(in mg)/Mass of solution (in Kg) ⇒ ppm = 20 / 10 = 2 So the ppm of lead in the sample is 2ppm.

Problem 8: What is the ppb of mercury in a sample that contains 0.01 g of mercury in 1000 L of air?

Given : mass of solute(in mg) = 0.01 g = 10,000 μg and Volution of solvent = 1000 L Mass of solution = Mass of water = 1000 L x 1 Kg/L = 1000 Kg (density of water is 1Kg/L or 1g/mL) ppb (parts per million) = Mass of solute(in μg)/Mass of solution (in Kg) ⇒ ppm = 10000 / 1000 = 10 So the ppm of lead in the sample is 10 ppb.

FAQs on Concentration of Solution

What is solubility, what is a dilute solution.

A dilute solution is solution which contains a smaller proportion of solute as compared to the proportion of solvent.

What is the difference between PPM and PPB?

PPM and PPB are both represents the concentration of very small scale and only difference between them is that PPM is 1000 times greater scale then PPB or PPB is 1000 times smaller scale then PPM i.e., 1 PPM = 1000 PPB

What is the difference between Molarity and Molality?

Molarity is the number of moles of solute per liter of solution, whereas molality is the number of moles of solute per kilogram of solvent.

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Mixture Problems With Solutions

Mixture problems and their solutions are presented along with their solutions. Percentages are also used to solve these types of problems.

Problem 1: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?

Solution to Problem 1: Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence x + 40 = y We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term. 20% x + 50% * 40 = 30% y Substitute y by x + 40 in the last equation to obtain. 20% x + 50% * 40 = 30% (x + 40) Change percentages into fractions. 20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100 Multiply all terms by 100 to simplify. 20 x + 50 * 40 = 30 x + 30 * 40 Solve for x. x = 80 liters 80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.

Problem 2: John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?

Solution to Problem 2: Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to make 100 ml. Hence x + y = 100 We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml. 2% x + 7% y = 5% 100 The first equation gives y = 100 - x. Substitute in the last equation to obtain 2% x + 7% (100 - x) = 5% 100 Multiply by 100 and simplify 2 x + 700 - 7 x = 5 * 100 Solve for x x = 40 ml Substitute x by 40 in the first equation to find y y = 100 - x = 60 ml

Problem 3: Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silver alloy?

Solution to Problem 3: Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence x + y =500 The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence 92.5% x + 90% y = 91% 500 Substitute y by 500 - x in the last equation to write 92.5% x + 90% (500 - x) = 91% 500 Simplify and solve 92.5 x + 45000 - 90 x = 45500 x = 200 grams. 200 grams of Sterling Silver is needed to make the 91% alloy.

Problem 4: How many Kilograms of Pure water is to be added to 100 Kilograms of a 30% saline solution to make it a 10% saline solution.

Solution to Problem 4: Let x be the weights, in Kilograms, of pure water to be added. Let y be the weight, in Kilograms, of the 10% solution. Hence x + 100 = y Let us now express the fact that the amount of salt in the pure water (which 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%. 0 + 30% 100 = 10% y Substitute y by x + 100 in the last equation and solve. 30% 100 = 10% (x + 100) Solve for x. x = 200 Kilograms.

Problem 5: A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?

Solution to Problem 5: The amount of the final mixture is given by 50 ml + 30 ml = 80 ml The amount of alcohol is equal to the amount of alcohol in pure water ( which is 0) plus the amount of alcohol in the 30% solution. Let x be the percentage of alcohol in the final solution. Hence 0 + 30% 50 ml = x (80) Solve for x x = 0.1817 = 18.75%

Problem 6: You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in terms of x. Find the ratio, in terms of x, of the alcohol in the final solution to the total amount of the solution. What do you think will happen if x is very large? Find x so that the final solution has a percentage of 15%.

Solution to Problem 6: Let us first find the amount of alcohol in the 10% solution of 200 ml. 200 * 10% = 20 ml The amount of alcohol in the x ml of 25% solution is given by 25% x = 0.25 x The total amount of alcohol in the final solution is given by 20 + 0.25 x The ratio of alcohol in the final solution to the total amount of the solution is given by [ ( 20 + 0.25 x ) / (x + 200)] If x becomes very large in the above formula for the ratio, then the ratio becomes close to 0.25 or 25% (The above function is a rational function and 0.25 is its horizontal asymptote). This means that if you increase the amount x of the 25% solution, this will dominate and the final solution will be very close to a 25% solution. To have a percentage of 15%, we need to have [ ( 20 + 0.25 x ) / (x + 200)] = 15% = 0.15 Solve the above equation for x 20 + 0.25 x = 0.15 * (x + 200) x = 100 ml

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Teens come up with trigonometry proof for Pythagorean Theorem, a problem that stumped math world for centuries

By Bill Whitaker

May 5, 2024 / 7:00 PM EDT / CBS News

As the school year ends, many students will be only too happy to see math classes in their rearview mirrors. It may seem to some of us non-mathematicians that geometry and trigonometry were created by the Greeks as a form of torture, so imagine our amazement when we heard two high school seniors had proved a mathematical puzzle that was thought to be impossible for 2,000 years. 

We met Calcea Johnson and Ne'Kiya Jackson at their all-girls Catholic high school in New Orleans. We expected to find two mathematical prodigies.

Instead, we found at St. Mary's Academy , all students are told their possibilities are boundless.

Come Mardi Gras season, New Orleans is alive with colorful parades, replete with floats, and beads, and high school marching bands.

In a city where uniqueness is celebrated, St. Mary's stands out – with young African American women playing trombones and tubas, twirling batons and dancing - doing it all, which defines St. Mary's, students told us.

Junior Christina Blazio says the school instills in them they have the ability to accomplish anything. 

Christina Blazio: That is kinda a standard here. So we aim very high - like, our aim is excellence for all students. 

The private Catholic elementary and high school sits behind the Sisters of the Holy Family Convent in New Orleans East. The academy was started by an African American nun for young Black women just after the Civil War. The church still supports the school with the help of alumni.

In December 2022, seniors Ne'Kiya Jackson and Calcea Johnson were working on a school-wide math contest that came with a cash prize.

Ne'Kiya Jackson: I was motivated because there was a monetary incentive.

Calcea Johnson: 'Cause I was like, "$500 is a lot of money. So I-- I would like to at least try."

Both were staring down the thorny bonus question.

Bill Whitaker: So tell me, what was this bonus question?

Calcea Johnson: It was to create a new proof of the Pythagorean Theorem. And it kind of gave you a few guidelines on how would you start a proof.

The seniors were familiar with the Pythagorean Theorem, a fundamental principle of geometry. You may remember it from high school: a² + b² = c². In plain English, when you know the length of two sides of a right triangle, you can figure out the length of the third.

Both had studied geometry and some trigonometry, and both told us math was not easy. What no one told  them  was there had been more than 300 documented proofs of the Pythagorean Theorem using algebra and geometry, but for 2,000 years a proof using trigonometry was thought to be impossible, … and that was the bonus question facing them.

Bill Whitaker: When you looked at the question did you think, "Boy, this is hard"?

Ne'Kiya Jackson: Yeah. 

Bill Whitaker: What motivated you to say, "Well, I'm going to try this"?

Calcea Johnson: I think I was like, "I started something. I need to finish it." 

Bill Whitaker: So you just kept on going.

Calcea Johnson: Yeah.

For two months that winter, they spent almost all their free time working on the proof.

CeCe Johnson: She was like, "Mom, this is a little bit too much."

CeCe and Cal Johnson are Calcea's parents.

CeCe Johnson:   So then I started looking at what she really was doing. And it was pages and pages and pages of, like, over 20 or 30 pages for this one problem.

Cal Johnson: Yeah, the garbage can was full of papers, which she would, you know, work out the problems and-- if that didn't work she would ball it up, throw it in the trash. 

Bill Whitaker: Did you look at the problem? 

Neliska Jackson is Ne'Kiya's mother.

Neliska Jackson: Personally I did not. 'Cause most of the time I don't understand what she's doing (laughter).

Michelle Blouin Williams: What if we did this, what if I write this? Does this help? ax² plus ….

Their math teacher, Michelle Blouin Williams, initiated the math contest.

Bill Whitaker: And did you think anyone would solve it?

Michelle Blouin Williams: Well, I wasn't necessarily looking for a solve. So, no, I didn't—

Bill Whitaker: What were you looking for?

Michelle Blouin Williams: I was just looking for some ingenuity, you know—

Calcea and Ne'Kiya delivered on that! They tried to explain their groundbreaking work to 60 Minutes. Calcea's proof is appropriately titled the Waffle Cone.

Calcea Johnson: So to start the proof, we start with just a regular right triangle where the angle in the corner is 90°. And the two angles are alpha and beta.

Bill Whitaker: Uh-huh

Calcea Johnson: So then what we do next is we draw a second congruent, which means they're equal in size. But then we start creating similar but smaller right triangles going in a pattern like this. And then it continues for infinity. And eventually it creates this larger waffle cone shape.

Calcea Johnson: Am I going a little too—

Bill Whitaker: You've been beyond me since the beginning. (laughter) 

Bill Whitaker: So how did you figure out the proof?

Ne'Kiya Jackson: Okay. So you have a right triangle, 90° angle, alpha and beta.

Bill Whitaker: Then what did you do?

Ne'Kiya Jackson: Okay, I have a right triangle inside of the circle. And I have a perpendicular bisector at OP to divide the triangle to make that small right triangle. And that's basically what I used for the proof. That's the proof.

Bill Whitaker: That's what I call amazing.

Ne'Kiya Jackson: Well, thank you.

There had been one other documented proof of the theorem using trigonometry by mathematician Jason Zimba in 2009 – one in 2,000 years. Now it seems Ne'Kiya and Calcea have joined perhaps the most exclusive club in mathematics. 

Bill Whitaker: So you both independently came up with proof that only used trigonometry.

Ne'Kiya Jackson: Yes.

Bill Whitaker: So are you math geniuses?

Calcea Johnson: I think that's a stretch. 

Bill Whitaker: If not genius, you're really smart at math.

Ne'Kiya Jackson: Not at all. (laugh) 

To document Calcea and Ne'Kiya's work, math teachers at St. Mary's submitted their proofs to an American Mathematical Society conference in Atlanta in March 2023.

Ne'Kiya Jackson: Well, our teacher approached us and was like, "Hey, you might be able to actually present this," I was like, "Are you joking?" But she wasn't. So we went. I got up there. We presented and it went well, and it blew up.

Bill Whitaker: It blew up.

Calcea Johnson: Yeah. 

Ne'Kiya Jackson: It blew up.

Bill Whitaker: Yeah. What was the blowup like?

Calcea Johnson: Insane, unexpected, crazy, honestly.

It took millenia to prove, but just a minute for word of their accomplishment to go around the world. They got a write-up in South Korea and a shout-out from former first lady Michelle Obama, a commendation from the governor and keys to the city of New Orleans. 

Bill Whitaker: Why do you think so many people found what you did to be so impressive?

Ne'Kiya Jackson: Probably because we're African American, one. And we're also women. So I think-- oh, and our age. Of course our ages probably played a big part.

Bill Whitaker: So you think people were surprised that young African American women, could do such a thing?

Calcea Johnson: Yeah, definitely.

Ne'Kiya Jackson: I'd like to actually be celebrated for what it is. Like, it's a great mathematical achievement.

Achievement, that's a word you hear often around St. Mary's academy. Calcea and Ne'Kiya follow a long line of barrier-breaking graduates. 

The late queen of Creole cooking, Leah Chase , was an alum. so was the first African-American female New Orleans police chief, Michelle Woodfork …

And judge for the Fifth Circuit Court of Appeals, Dana Douglas. Math teacher Michelle Blouin Williams told us Calcea and Ne'Kiya are typical St. Mary's students.  

Bill Whitaker: They're not unicorns.

Michelle Blouin Williams: Oh, no no. If they are unicorns, then every single lady that has matriculated through this school is a beautiful, Black unicorn.

Pamela Rogers: You're good?

Pamela Rogers, St. Mary's president and interim principal, told us the students hear that message from the moment they walk in the door.

Pamela Rogers: We believe all students can succeed, all students can learn. It does not matter the environment that you live in. 

Bill Whitaker: So when word went out that two of your students had solved this almost impossible math problem, were they universally applauded?

Pamela Rogers: In this community, they were greatly applauded. Across the country, there were many naysayers.

Bill Whitaker: What were they saying?

Pamela Rogers: They were saying, "Oh, they could not have done it. African Americans don't have the brains to do it." Of course, we sheltered our girls from that. But we absolutely did not expect it to come in the volume that it came.  

Bill Whitaker: And after such a wonderful achievement.

Pamela Rogers: People-- have a vision of who can be successful. And-- to some people, it is not always an African American female. And to us, it's always an African American female.

Gloria Ladson-Billings: What we know is when teachers lay out some expectations that say, "You can do this," kids will work as hard as they can to do it.

Gloria Ladson-Billings, professor emeritus at the University of Wisconsin, has studied how best to teach African American students. She told us an encouraging teacher can change a life.

Bill Whitaker: And what's the difference, say, between having a teacher like that and a whole school dedicated to the excellence of these students?

Gloria Ladson-Billings: So a whole school is almost like being in Heaven. 

Bill Whitaker: What do you mean by that?

Gloria Ladson-Billings: Many of our young people have their ceilings lowered, that somewhere around fourth or fifth grade, their thoughts are, "I'm not going to be anything special." What I think is probably happening at St. Mary's is young women come in as, perhaps, ninth graders and are told, "Here's what we expect to happen. And here's how we're going to help you get there."

At St. Mary's, half the students get scholarships, subsidized by fundraising to defray the $8,000 a year tuition. Here, there's no test to get in, but expectations are high and rules are strict: no cellphones, modest skirts, hair must be its natural color.

Students Rayah Siddiq, Summer Forde, Carissa Washington, Tatum Williams and Christina Blazio told us they appreciate the rules and rigor.

Rayah Siddiq: Especially the standards that they set for us. They're very high. And I don't think that's ever going to change.

Bill Whitaker: So is there a heart, a philosophy, an essence to St. Mary's?

Summer Forde: The sisterhood—

Carissa Washington: Sisterhood.

Tatum Williams: Sisterhood.

Bill Whitaker: The sisterhood?

Voices: Yes.

Bill Whitaker: And you don't mean the nuns. You mean-- (laughter)

Christina Blazio: I mean, yeah. The community—

Bill Whitaker: So when you're here, there's just no question that you're going to go on to college.

Rayah Siddiq: College is all they talk about. (laughter) 

Pamela Rogers: … and Arizona State University (Cheering)

Principal Rogers announces to her 615 students the colleges where every senior has been accepted.

Bill Whitaker: So for 17 years, you've had a 100% graduation rate—

Pamela Rogers: Yes.

Bill Whitaker: --and a 100% college acceptance rate?

Pamela Rogers: That's correct.

Last year when Ne'Kiya and Calcea graduated, all their classmates went to college and got scholarships. Ne'Kiya got a full ride to the pharmacy school at Xavier University in New Orleans. Calcea, the class valedictorian, is studying environmental engineering at Louisiana State University.

Bill Whitaker: So wait a minute. Neither one of you is going to pursue a career in math?

Both: No. (laugh)

Calcea Johnson: I may take up a minor in math. But I don't want that to be my job job.

Ne'Kiya Jackson: Yeah. People might expect too much out of me if (laugh) I become a mathematician. (laugh)

But math is not completely in their rear-view mirrors. This spring they submitted their high school proofs for final peer review and publication … and are still working on further proofs of the Pythagorean Theorem. Since their first two …

Calcea Johnson: We found five. And then we found a general format that could potentially produce at least five additional proofs.

Bill Whitaker: And you're not math geniuses?

Bill Whitaker: I'm not buying it. (laughs)

Produced by Sara Kuzmarov. Associate producer, Mariah B. Campbell. Edited by Daniel J. Glucksman.

Bill Whitaker is an award-winning journalist and 60 Minutes correspondent who has covered major news stories, domestically and across the globe, for more than four decades with CBS News.

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AI’s Trust Problem

• Bhaskar Chakravorti

As AI becomes more powerful, it faces a major trust problem. Consider 12 leading concerns: disinformation, safety and security, the black box problem, ethical concerns, bias, instability, hallucinations in LLMs, unknown unknowns, potential job losses and social inequalities, environmental impact, industry concentration, and state overreach. Each of these issues is complex — and not easy to solve. But there is one consistent approach to addressing the trust gap: training, empowering, and including humans to manage AI tools.

Twelve persistent risks of AI that are driving skepticism.

With tens of billions invested in AI last year and leading players such as OpenAI looking for trillions more, the tech industry is racing to add to the pileup of generative AI models. The goal is to steadily demonstrate better performance and, in doing so, close the gap between what humans can do and what can be accomplished with AI.

• Bhaskar Chakravorti is the Dean of Global Business at The Fletcher School at Tufts University and founding Executive Director of Fletcher’s Institute for Business in the Global Context . He is the author of The Slow Pace of Fast Change .

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AI, EVs, and satellites are tackling the climate crisis. But they have environmental downsides. This hour, TED speakers explain how to use these tools without making global warming worse.

Guests include AI researchers Sasha Luccioni and Sims Witherspoon, climate researcher Elsa Dominish and astrodynamicist Moriba Jah.

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New Journal of Chemistry

Study on the removal of so 4 2− and ca 2+ from potassium chloride brine via a method combining calcium chloride and carbon dioxide.

* Corresponding authors

a National Fundamental Research Laboratory of New Hazardous Chemicals Assessment and Accident Analysis, Institute of Applied Electrochemistry, Beijing University of Chemical Technology, Beijing 100029, China E-mail: [email protected] Fax: +8610-64435452 Tel: +8610-64435452

b State Key Laboratory of Chemical Resource Engineering, Beijing University of Chemical Technology, Beijing 100029, China E-mail: [email protected]

Purification treatment is needed to control the concentration of SO 4 2− and Ca 2+ in brine to solve the problem of low electrolysis efficiency for the electrolysis of KCl solution in the chlor-alkali industry. However, the traditional barium method to remove SO 4 2− faces the problem of high cost and environmental pollution. Herein, inexpensive and environmentally friendly anhydrous calcium chloride was used as the precipitant to replace barium chloride. The conditions for SO 4 2− removal at room temperature were optimized by studying the precipitant doping ratio, reaction time, and stirring speed. The results indicated that the concentration of SO 4 2− in brine could be reduced to less than 5 g L −1 at a raw material ratio of n Ca 2+ to n SO 4 2− of 1.1 : 1, a stirring speed of 300 rpm, and a reaction time of 120 min. Ca 2+ in the solution could be removed by introducing carbon dioxide, and the removal rate of Ca 2+ could reach 99.89% by adjusting the pH of the solution to an appropriate value. This work paves a new way for efficient brine refining to boost sustainable development in the chlor-alkali industry.

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H. Ren, Q. Wang, Y. Sun, Y. Chen, P. Wan and J. Pan, New J. Chem. , 2024, Advance Article , DOI: 10.1039/D4NJ01044B

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Esade Entrepreneurship Institute

"99% of start-ups actually fail, so you have to think that you are most likely going to fail".

Listen to this podcast via  Spotify  /  Apple Podcasts  /  Google Podcasts

The story of Polaroo, a utility management platform, is an inspiring tale of perseverance, problem-solving, and the power of strong relationships. This is why Davide Rovera, Manager of eWorks, interviewed the co-founders Marc and David Rovira (Esade graduate) at the Esade Doers podcast series, about their entrepreneurial journey, the challenges they faced and their vision and commitment to solving pressing problems.

>> Check the full interview here: https://dobetter.esade.edu/en/entrepreneurs-polaroo

Polaroo, founded by brothers David and Marc Rovira, is a groundbreaking startup that is revolutionizing how businesses manage services and payments. With their expertise in Business Law and Engineering, they have developed a service app that offers end-to-end management and payment control for B2Bs. By centralizing services, bills, and SaaS subscriptions into a single platform, Polaroo streamlines processes, optimizes costs through bundling and bulk-buying, and provides financial stability by consolidating payments into a single monthly transaction. The Rovira brothers' unwavering entrepreneurial spirit and shared vision have propelled Polaroo to new heights, offering businesses an innovative solution to enhance efficiency and reduce unnecessary expenses.

Behind Polaroo lies an inspiring journey of determination, sacrifice, and the power of collaboration. David and Marc's entrepreneurial drive was sparked at a young age, organizing events and finding ways to turn their passions into profitable ventures.

Starting their entrepreneurship journey

For Marc and David, the decision to become entrepreneurs was not sudden but rather a culmination of experiences and a deep-rooted desire to make a difference. Through their journey, they have learned the value of a strong team, embracing challenges head-on, and never shying away from taking calculated risks. Polaroo serves as a testament to the power of innovation and the impact that dedicated individuals can have on transforming industries. Marc's frustration with managing recurrent expenses while living in different apartments sparked the initial idea. Recognizing the complexities and inefficiencies in dealing with various utility providers, they saw an opportunity to bridge the gap between consumers and these recurring expenses.

The initial focus of Polaroo was on helping consumers manage their utility payments more efficiently. However, as the founders delved deeper into the concept, they recognized that, in a world increasingly dominated by the subscription economy, the problem extended beyond utilities. Subscriptions, insurance, and other recurring expenses were also part of the equation. They saw an opportunity to address this broader issue and provide a comprehensive solution to simplify the management of all recurring expenses.

Pivoting is key to finding your best market-fit

Despite lacking prior experience in the industry, Marc and David leveraged their problem-solving skills and outsider perspective to tackle the challenges of the energy and utility sectors. Their fresh approach allowed them to identify innovative solutions that were often overlooked by industry experts, mired in the complexities of their respective fields. They saw the problem from a financial standpoint, focusing on the consumer experience rather than the technical intricacies. By combining their expertise and continuously seeking feedback from industry specialists, they navigated the learning curve and successfully built a company that disrupts the traditional utility management landscape.

As they continued their journey, they observed that businesses were using their software more than individual consumers. They realized that businesses faced a monthly obligation to manage their expenses, making their solution even more critical in the B2B realm. This organic shift led them to pivot and focus on serving businesses rather than individual consumers. Businesses were particularly keen on effectively managing their expenses and cash flow, and by catering to businesses' ongoing accounting needs, Polaroo positioned itself as a valuable tool for financial management in the subscription-based economy and enabled them to tap into a rapidly expanding market and set the stage for its rapid growth as a fintech company.

Tips and learnings when facing challenges to build a sustainable-growing startup

One of the key lessons Marc and David learned early on was the importance of product-market fit. They realized that instead of pushing people to understand the problem their product aimed to solve, it was more effective to create a solution that resonated with users. By focusing on building a product that addressed a real need and allowing users to feel the value intuitively, Polaroo gained a stronger position in the market. This approach validated their product-market fit and showcased the potential for future growth.

In the initial stages, the co-founders dedicated time to a learning phase, drawing on their prior experiences and expertise. They understood the significance of a well-constructed minimum viable product (MVP). Contrary to the misconception that an MVP should be complex, Marc and David opted for an advanced Excel file to manage and optimize recurrent payments—an uncomplicated yet efficient solution. This approach allowed them to quickly test and iterate, gathering valuable feedback to refine their offering.

Marc emphasized the importance of doing things that don't scale, a concept advocated by Paul Graham, one of the founders of Y Combinator. He stressed that these non-scalable actions are not meant to become scalable but rather serve as a litmus test for product demand. By focusing on delivering a solution that solves a problem, even through manual or non-scalable means, entrepreneurs can gauge whether there is genuine interest in their product. Whether it's an Excel file or an AI-powered app, what matters most is the ability to address the core problem in the simplest, quickest, and most cost-effective way.

The power of the MVP became evident to the Polaroo team. The Excel file they built, though lacking the fancy aesthetics of a polished app, effectively solved the problem at hand. Users perceived it as a powerful tool, oblivious to the simple backend operations. Marc explained the importance of distinguishing between how the solution is perceived by the user and the technical complexities happening behind the scenes. By testing and validating the functionality, entrepreneurs can determine whether a feature works before investing resources to scale it.

Reflecting on their journey, the co-founders emphasized the strength of their bond as brothers. Having been partners throughout their lives, they developed an unwavering trust and respect for each other. The shared experiences, challenges, and multiple connections they built over time created a solid foundation for their partnership. Marc and David recognized that their relationship extended beyond work and encompassed various facets of life. This multifaceted connection allowed them to weather disagreements, maintain a strong support system, and reduce stress levels. Their example challenges the notion that starting a company with close family members is inherently risky, as evidenced by the success of other sibling-led companies like Stripe.

With a team of 15 dedicated individuals working full-time and over 20 collaborators, Polaroo has grown steadily. The company's focus on building trust, maintaining a problem-solving mindset, and nurturing relationships have contributed to its success. Even if the company were to face failure, Marc and David's resilient spirit and shared experiences would provide the foundation for future adventures.

As the interview came to a close, the co-founders shared the story behind the name "Polaroo." Inspired by their early days working in a cold garage, they began referring to it as their "igloo.".

As Polaroo continues to grow and evolve, it remains committed to providing individuals and businesses with the tools to effectively manage their recurring expenses. The entrepreneurial journey of Marc and his team exemplifies the transformative power of determination, problem-solving, and a relentless pursuit of one's vision.

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