c
The vapour pressure of pure \(CHCl _{3}\) and \(CH _{2} Cl _{2}\) are \(200\) and
\(41.5\, atm\) respectively. The weight of \(CHCl _{3}\) and \(CH _{2} Cl _{2}\) are respectively \(11.9 \,g\) and \(17\, gm\).

The molar mass of \(CHCl _{3}\), is \(119.5 \,g / mol\) and the molar mass of \(CH _{2} Cl _{2}\) is \(85 \,g / mol\)

The number of moles of \(CHCl _{3}\) is calculated below.

\(n _{ CHCl _{3}} =\frac{11.9}{119.5 \,g / mol } \)

\(=0.1\)

The number of moles of \(CH _{2} Cl _{2}\) is calculated below.

\(n _{ CH _{2} Cl _{2}}=\frac{17 g }{85 g / mol }\)

\(=0.2\) 

So, total moles are \(01 .+0.2=0.3\).

According to the Raoults' law, the vapour pressure of solution is

calculated below.

\(P _{ T }= P _{ CHCl _{3}}^{\circ} \times x _{ CHCl _{3}}+ P _{ CH _{2} Cl _{2}}^{\circ} \times X _{ CH _{2} Cl _{2}}\)

\(=200 \times \frac{0.1}{0.3}+41.5 \times \frac{0.2}{0.3}\)

\(=94.3\) \(atm\)