Question

what is the vapor pressure of water above a solution prepared by dissolving 28.5 g glycerin (C_(3)H_(8)O_(3)) in 0.125 kg water at 243 K. The vapor pressure of pure water at that temperature is 233.70 torr.

Answer 1

From Raoult's law, $P_{solution}=X_{solvent}×P_{solvent}$

Where denotes vapor pressure, and X represent mole fraction.

$Mole fraction =\frac{moles of solvent}{moles of solute+moles of solvent}$

$Vapor pressure=\frac{\frac{0.125×1000}{18}}{\frac{28.5}{92}+\frac{0.125×1000}{18}}×233.7=223.72$