Question

Vapour pressure of chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$ at $25^{\circ }C$ are $200mm$ of $Hg$ and $41.5mm$ of $Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5g$ of $CHC{l}_3$ and $40g$ of $C{H}_{2}C{l}_2$ at the same temperature will be (Molecular mass of $CHC{l}_3=119.5u$ and molecular mass of $C{H}_{2}C{l}_2=85u$ )

• A. 615.0 mm Hg
• B. 347.9 mm Hg
• C. 285.5 mm Hg
• D. 90.38 mm Hg

Answer 1

Answer: (B)

Molar mass of $C{H}_{2}C{l}_2=12\times 1+1\times 2+35.5\times 2=85gmo{l}^{-1}$
Molar mass of $CHC{l}_3=12\times 1+1\times 1+35.5\times 3=119.5gmo{l}^{-1}$
Moles of $C{H}_{2}C{l}_2=40g/85gmo{l}^{-1}=0.47mol$
Moles of $CHC{l}_3=25.5g/119.5gmo{l}^{-1}=0.213mol$
Total number of moles $=0.47+0.213=0.683mol$
Mole fraction of component 2
$<br/>=0.47mol/0.683mol=0.688<br/>$
Mole fraction of component 1
$<br/>=1.00-0.688=0.312<br/>$
We know that:
$<br/>P_T=p_1^0+\left(p_2^0-p_1^0\right)x_2<br/>$
$<br/>=200+(415-200)\times 0.688<br/>$
$<br/>=200+147.9<br/>$
$<br/>=347.9mmHg<br/>$