Answer
$P_{total}=92.8\,atm$
$P_{CHCl_{3}}=24.3\,atm$
Work Step by Step
$n(CHCl_{3})=\frac{3.23\,g}{119.38\,g/mol}=0.02705646\,mol$
$n(CH_{4})=\frac{1.22\,g}{16.04\,g/mol}=0.07606\,mol$
$R=0.0821\,L\,atm\,mol^{-1}K^{-1}$
$T=(275+273)K=548\,K$
$V=50.0\,mL=0.0500\,L$
$P_{total}= \frac{n_{total}RT}{V}$
$=\frac{[n(CHCl_{3})+n(CH_{4})]RT}{V}$
$=\frac{(0.02705646\,mol+0.07606\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(548\,K)}{0.0500\,L}$
$=92.8\,atm$
$P_{CHCl_{3}}= \frac{n_{CHCl_{3}}RT}{V}$
$=\frac{(0.02705646\,mol)(0.0821\,L\,atm\,mol^{-1}K^{-1})(548\,K)}{0.0500\,L}$
$=24.3\,atm$
