# Question 66b16

May 22, 2017

$\text{150 g}$

#### Explanation:

Assuming that you're dealing with a non-volatile solute, you can say that the vapor pressure of the solution, ${P}_{\text{sol}}$, will depend on the mole fraction of the solvent, ${\chi}_{\text{water}}$, and on the vapor pressure of the pure solvent, ${P}_{\text{water}}^{\circ}$. as given by Raoult's Law

color(blue)(ul(color(black)(P_"sol" = chi_"water" * P_"water"^@)))" "color(darkorange)("(*)")

Now, the mole fraction of water is defined as the ratio between the number of moles of water and the total number of moles present in the solution.

If you take $m$ to be the mass of the solute in grams, you can say that the solution will contain

m color(red)(cancel(color(black)("g"))) * "1 mole solute"/(60color(red)(cancel(color(black)("g")))) = (m/60) $\text{moles solute}$

and

180 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "9.992 moles H"_2"O"

The total number of moles in the solution will be equal to

$\left(\frac{m}{60}\right) \textcolor{w h i t e}{.} \text{moles" + "9.992 moles} = \left(\frac{m + 599.52}{60}\right)$ $\text{moles}$

The mole fraction of water will thus be

chi_"water" = (9.992 color(red)(cancel(color(black)("moles"))))/(( (m + 599.52)/60)color(red)(cancel(color(black)("moles")))) = 599.52/(m + 599.52)#

You know that the vapor pressure of the solution must be $\frac{4}{5} \text{th}$ the vapor pressure of the pure solvent, so

${P}_{\text{sol" = 4/5 * P_"water}}^{\circ}$

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to get

$\frac{4}{5} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{P}_{\text{water"^@))) = chi_"water" * color(red)(cancel(color(black)(P_"water}}^{\circ}}}}$

This is equivalent to

$\frac{599.52}{m + 599.52} = \frac{4}{5}$

Solve for $m$ to find

$4 \cdot \left(m + 599.52\right) = 599.52 \cdot 5$

$m = \frac{599.52 \cdot 5 - 599.52 \cdot 4}{4}$

$m = \frac{599.52}{4} = 149.88$

Therefore, you can say that the mass of the solute is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass solute = 150 g}}}}$

I'll leave the answer rounded to two sig figs.