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Sucrose is a non volatile, non-ionizing solute in water.

Determine the vapor pressure lowering, at 27 degrees Celsius, of a solution of 75.0 grams sucrose, C12H22O11 dissolved in 180 grams of water.

Note: the vapor pressure of water at 27 degrees Celsius is 26.7 torr.

Verified Answer :

Edufirst

When a non volatile solute is added to a solvent, the vapor pressure of the solvent will decrease.

 Vapor pressure is a colligative property. Which means that it will be affected by the number of particles of solute.

The relationship between the concentration of the solute and the vapor pressure is given by Roult's Law

ΔP = Xsolute P, where ΔP is vapor pressure lowering, Xsolute is the molar fractionof the solute, and P is the vapor pressure of the solvent.

Then lets find Xsolute

X solute = # moles solute / # moles solution

# moles solute = 75.0 grams / molar mass

molar mass of C12H22O11 = 1212g/mol + 221g/mol + 1116g/mol = 342 g/mol

# moles solute = 75.0/342 mol = 0.219 mol

# moles of solvent = grams of solvent / molar mass of solvent

# moles of solvent = 180 grams /18 g/mol = 10 mols

Xsolute = 0.219 mol solute / (10 + 0.219) mol solution =0.0215

ΔP = 0.0215*26.7 torr = 0.57 torr

Answer= 0.57 torr

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