Question

What is the vapor pressure of a 45.0 % solution of glucose, c6h12o6, at 90.0°c, given that the vapor pressure of pure water at that temperature is 526 mm hg?

Answer 1

Weight of solute = 45 g and weight of solvent = 55 g
M.wt of solute = 180      P⁰ = 526 mmHg
number of moles of solute = 45 / 180 = 0.25 mol
number of moles of solvent = 55 / 18 = 3.05 mol
According to Rault's law:
$\frac{P^{0} - P^{s} }{P^{0} } = \frac{n_{solute} }{n_{solute} + n_{solvent}}$ = 0.25 / (0.25 + 3.05)

$\frac{526 - P^{s} }{526} = 0.076$

$P^{s} = 486.15 mmHg$