Ways of Expressing Concentrations of Solutions Mass Percentage Mass % of A = mass of A in solution...

Ways of Expressing Ways of Expressing Concentrations of Concentrations of

SolutionsSolutions

Mass Percentage

Mass % of A =mass of A in solutiontotal mass of solution 100

Parts per Million andParts per Billion

ppm =mass of A in solutiontotal mass of solution 106

Parts per Million (ppm)

Parts per Billion (ppb)

ppb =mass of A in solutiontotal mass of solution 109

moles of Atotal moles in solutionXA =

Mole Fraction (X)

• In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

mol of soluteL of solutionM =

Molarity (M)

• Because volume is temperature dependent, molarity can change with temperature.

mol of solutekg of solventm =

Molality (m)

Because neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependent.

SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations

(a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 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 5.4g 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 EXERCISEA 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 solutionsPositive 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 > pB

0XB

∴ pA + pB > pA0XA + pB

0XB  

1.Do not obey Raoult’s law.    

2.Hmix<0. Exothermic dissolution; heat is evolved.  

3.Vmix <0. Volume is decreased during dissolution.    

4.pA < pA0XA; pB < pB

0XB

∴ pA + pB < pA0XA + pB

0XB    

Ideal solutions Non-ideal solutionsPositive 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.