#### Answer

Partial pressure of acetone= 172 torr.
Partial pressure of chloroform= 148 torr.

#### Work Step by Step

Let $p_{1}$ be the partial pressure of acetone.
Given $p_{1}^{0}$= 345 torr and mole fraction, $x_{1}=\frac{n_{1}}{n_{1}+n_{2}}= \frac{0.550 mol}{(0.550+0.550)mol}= 0.5$
Then, using Raoult's law, we have
$p_{1}=p_{1}^{0}x_{1}= 345 torr\times0.5\approx 172$ torr.
Similarly, let $p_{2}$ be the partial pressure of chloroform.
$p_{2}= p_{2}^{0}x_{2}= 295 torr\times \frac{n_{2}}{n_{1}+n_{2}}$= $295 torr \times0.5\approx 148 torr$.