Question

1) A solution of 1.01 g of benzoic acid in 8.02 g of lauric acid had a freezing point of 38.9 °C. What is the molar mass of benzoic acid? The freezing point of lauric acid is 44.0 °C and Kf is 3.60 °C.kg/mol.

2) Calculate the osmotic pressure of a 0.150 M aqueous solution of caalcium chloride at 25°C.

Answer 1

part A

Tf = Kf x m x i

where

Tf = 44-38.9 = 5.1 ºC

Kf = 3.6 ºC kg / mol

i = vant haf factor which is 1 for lauric acid

plug in these values in above equation

5.1 ºC = 3.6 ºC kg / mol x m x 1

m = 5.1 ºC / 3.6 ºC kg / mol = 1.4167 mol / kg

now use the molality formula

molality (m) = (mass / molar mass) x 1000 / mass of solvent

1.4167 mol / kg = (1.01 g / molar mass ) x 1000 / 8.02 g

molar mass = 88.9 g/mol

Part B

osmatic pressure

=M x R x T x i

where

= osmatic pressure

M = molarity = 0.15 M

T = temperature = 273+25 = 298 K

R = gas constant = 0.0821 L atm / mol K

i = vant haf factor = 3 for CaCl2

because CaCl2 can ionize and give the one Ca2+ and 2Cl-

plug in these values in above equation

= 0.15 mol / L x 0.082 L atm / mol K x 298 K x 3

= 11.00 atm