Question

"Vapour pressure of chloroform and dichloromethane at $298\, K$ are $200\, mm$ of $Hg$ and $415\, mm$ of $Hg$ respectively".
What is the vapour pressure of solution prepared by mixing $25.5\, g$ of $CHCl _{3}$ and $40\, g$ of $CH _{2} Cl _{2}$ at $298 K$ ?

• A. $347.9\, mmHg$
• B. $300\, mmHg$
• C. $200\, mmHg$
• D. $147.9\, mmHg$

Answer 1

Answer: (A)

Molar mass of $CH _{2} Cl _{2}=12 \times 1+1 \times 2+35.5 \times 2$
$=85\, g\, mol ^{-1}$
Molar mass of $CHCl _{3} =12 \times 1+1 \times 1+35.5 \times 3$
$=119.5\, g\, mol ^{-1}$
Moles of $CHCl _{3}=\frac{25.5}{119.5}=0.213\, mol$
Moles of $CH _{2} Cl _{2}=\frac{40 g }{85 g mol ^{-1}}=0.47\, mol$
Total number of moles $=0.47+0.213=0.683\, mol$
$x_{ CH _{2} Cl _{2}} =\frac{0.47 mol }{0.683 mol }=0.688$
$x_{ CHCl _{2}} =1.00-0.688=0.312$
Using equation, $p_{\text {total }} =p_{1}^{\circ}+\left(p_{2}^{\circ}-p_{1}^{\circ}\right) x_{2}$
$=200+(415-200) \times 0.688$
$=200+147.9$
$=347.9\, mmHg$