Question

Calculate the vapor pressure depression of a solution that 218 gram glucose (180.2 g/mol) is dissolved in 460 mL of water (18.01 g/mol). The temperature of the solution is 30°C. Assume the density of solution and water as 1.00 g/mL, the vapor pressure of pure water at 30°C 31.82 mm Hg.

Answer 1

Given, vapour pressure of pure water at 30°C

P1° = 31.82 mm Hg

Molecular mass of water= 18.01 g/mol

Volume of water= 460 mL

Census density of water is 1 gram/mL,

Mass of water = 460 gram

Moles of water = mass / molecular mass

Moles of water (n1) = 460 / 18.01

Moles of water (n1) = 25.54 mols

Molecular mass of glucose= 180.2 g/mol

Mass of glucose= 218 g

Moles of glucose = mass / molecular mass

Moles of glucose(n2) = 218 / 180.2

Moles of glucose (n2) = 1.21 mols

Lowering in vapour pressure can express by following formula-

(P1°- P1)/P1° = n2/n1

∆P / 31.82 = 1.21 / 25.54

∆P = 0.0474 * 31.82

∆P = 1.507

lowering in vapour pressure should be 1.507 mm Hg.