Question

1) What will the solution vapor pressure be for a solution of 11.07 mol of sucrose (C₁₂H₂₂O₁₁) dissolved in 1.000L of water at 20.0 °C? The vapor pressure of pure water at 20.0 °C is 17.5 torr. (Show work)

2) Calculate the freezing point of a solution containing 0.43 of sucrose in 150 mL of water. K_f = 1.86 °C/m (Show Work)

Answer 1

Moles of sucrose = 11.07 mol

Moles of water = 1000 g. / 18.0 g/mol = 55.55 mol

Mole fraction of sucrose = 11.07 / (11.07+55.55) = 0.166

we know that,

Relative lowering of vapour pressure = mole fraction of solute

(p0 - p) / p0 = mole fraction of sucrose

(17.5 - p) / 17.5 = 0.166

17.5 - p = 2.91

p = vapour pressure of solution = 14.59 torr

(2)

Mass of sucrose = 0.43 g.

Molar mass of sucrose = 342 g/mol

Moles of sucrose = mass / molar mass = 0.43 / 342 = 0.00126 mol

Mass of water = 150 g. = 0.150 kg.

Molality of sucrose = moles of sucrose / mass of water in kg

m = 0.00126 / 0.150

m = 0.00838 m

We that depreesion in freezing point = K_f * m

0.00 - T_f = 1.86 * 0.00838

T_f = - 0.0156 ⁰C