Calculate the vapor pressure of water above a solution prepared by dissolving 28.5 g of glycerin (c3h8o3) in 135 g of water at 343 k. (the vapor pressure of water at 343 k is 233.7 torr.)

Data:

Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.

Quesiton: Vapor pressure of water

Solution:

Raoult's Law: The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature  multiplied by its mole fraction.

Formula: p = Xsolvent * P pure solvent

X solvent = moles solvent / moles of solution

molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol

moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol

molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol

moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol

moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol

Xsolvent = 7.50mol / 7.81mol = 0.960

p = 233.7 torr * 0.960 = 224.4 torr

224.4 torr
 

Igonna be honest i dunno that one srry