Calculate the freezing point and boiling point of a solution that contains 65.6 g NaCl and 42.4 g KBr dissolved in 750.3 mL H2O.

Answer 1

For freezing point:

∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 2 for NaCl and 2 for KBr = 4 for the mixture

m = molality = mols solute / kg solvent = see below for calculation of m

K = freezing constant for water = 1.86 º/m

calculation of m (molality)

molar mass NaCl = 58.4 g / mol; 65.6 g NaCl x 1 mol / 58.4 g = 1.12 mols NaCl

molar mass KBr = 119 g / mol; 42.4 g KBr x 1 mol / 119 g = 0.356 mols KBr

Total moles added = 1.12 + 0.356 = 1.476 mols

Kg of water = 750.3 mls x 1 g / ml = 750.3 g = 0.7503 kg (assuming a density of 1 g / ml)

molality (m) = 1.476 mols / 0.7503 kg = 1.967 m

Calculation of freezing point:

∆T = (4)(1.967)(1.86) = 14.6ºC

New freezing point of solution = -14.6ºC

For boiling point:

∆T = imK

∆T = change inboiling point = ?

i = van't Hoff factor = 2 for NaCl and 2 for KBr = 4 for the mixture

m = molality = mols solute / kg solvent = 1.967 m (see above calculations in part a)

K = boiling point constant for water = 0.512º/m

Calculation of boiling point:

∆T = (4)(1.967)(0.512) = 4.03º

New boiling point of solution = 100º + 4.03º = 104.3º