# Ways of Expressing Concentrations of Solutions

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Ways of Expressing Concentrations of Solutions

Mass Percentage mass of A in solution Mass % of A =
total mass of solution Mass % of A =  100

Parts per Million and Parts per Billion
Parts per Million (ppm) mass of A in solution total mass of solution ppm =  106 Parts per Billion (ppb) mass of A in solution total mass of solution ppb =  109

total moles in solution
Mole Fraction (X) moles of A total moles in solution XA = In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

Molarity (M) mol of solute M = L of solution
Because volume is temperature dependent, molarity can change with temperature.

Molality (m) mol of solute m = kg of solvent
Because neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependent.

Changing Molarity to Molality
If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4g of Zn2+ What is the concentration of Zn2+ in parts per million? PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? PRACTICE EXERCISE A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution.

Ideal solutions Non-ideal solutions Positive deviation from Raoult’s law Negative deviation from Raoult’s law 1.Obey Raoult’s law at every range of concentration.   2.Hmix = 0; neither is evolved nor absorbed during dissolution.   3.Vmix = 0; total volume of solution is equal to sum of volumes of the components.   4.P = pA + pB = pA0XA + pB0XB    i.e., pA =     1.Do not obey Raoult’s law.     2.Hmix>0. Endothermic dissolution; heat is absorbed.   3.Vmix > 0. Volume is increased after dissolution. 4.pA > pA0XA; pB > pB0XB ∴ pA + pB > pA0XA + pB0XB   2.Hmix<0. Exothermic dissolution; heat is evolved.   3.Vmix <0. Volume is decreased during dissolution.     4.pA < pA0XA; pB < pB0XB ∴ pA + pB < pA0XA + pB0XB

Ideal solutions Non-ideal solutions Positive deviation from Raoult’s law Negative deviation from Raoult’s law     5.A—A, A—B, B—B interactions should be same, i.e., ‘A’ and ‘B’ are identical in shape, size and character.       6. Escaping tendency of ‘A’ and ‘B’ should be same in pure liquids and in the solution.   Examples: dilute solutions; benzene + toluence: n-hexane + n-heptane; chlorobenzene + bromobenzene; n-butyl chloride + n-butyl bromide. 5.A—B attractive force should be weaker than A—A and B—B attractive forces. ‘A’ and ‘B’ have different shape, size and character.       6. ‘A’ and B’ escape easily showing higher vapour pressure than the expected value.     Examples: acetone + ethanol acetone + CS2; water + methanol; water + ethanol; CCl4 + toluene; CCl4 + CHCl3; acetone + benzene; CCl4 + CH3OH; Cyclohexane + ethanol 5. A—B attractive force should be greater than A—A and B—B attractive forces. ‘A’ and ‘B’ have different shape, size and character.   6. Escaping tendency of both components ‘A’ and ‘B’ is lowered showing lower vapour pressure than expected ideally.   Examples: acetone + aniline; acetone + chloroform; CH3OH + CH3COOH; H2O + HNO3; Choloroform + diethyl ether, water + HCl; acetic acid + pyridine; chloroform + benzene.

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4g of Zn2+ What is the concentration of Zn2+ in parts per million? Solution Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan: We can calculate the mass percentage by using Equation The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve:  Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%. (b) Analyze: In this case we are given the number of micrograms of solute. Because 1g is 1  10–6 g, 5.4g = 5.4  10–6 g. Plan: We calculate the parts per million using Equation 13.6. Solve:   PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? Answers: (a) 2.91%, (b) 90.5 g of NaOCl Answers: (a) m, (b) 9.00  10–3