Question

If an ideal solution contains 3.50 mol of a non-volatile solute and 15.8 mol of solvent, and it has a vapor pressure of 22.6 torr, what is the vapor pressure of the pure solvent (in torr)

Answer 1

Answer: The vapor pressure of the solution is 18.5 torr

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=i\times x_2$

where,

$\frac{p^o-p_s}{p^o}$= relative lowering in vapor pressure

i = Van'T Hoff factor = 1 (for non electrolytes)

$x_2$ = mole fraction of solute

Given :  moles of solute = 3.50

moles of solvent = 15.8

Total moles = moles of solute  + moles of solvent = 3.50 + 15.8 = 19.3

$x_2$ = mole fraction of solute = $\frac{3.50}{19.3}=0.181$

$\frac{22.6-p_s}{22.6}=1\times 0.181$

$p_s=18.5torr$

Thus the vapor pressure of the solution is 18.5 torr

Answer 2

Answer:

27.6 torr

Explanation:

First, find the mole fraction of the solvent.

Xsolvent=15.8 mol(15.8 mol+3.50 mol)=0.81865

Next, use Rault's Law and solve for the vapor pressure of the pure solvent (P∘solvent).

PsolutionP∘solvent=XsolventP∘solvent=PsolutionXsolvent=22.6torr0.81865=27.606torr

The answer has three significant figures, so round to 27.6torr.