Question

Vapour pressure of chloroform $\left(CHC{l}_3\right)$ and dichloromethane $\left(C{H}_{2}C{l}_2\right)$ at ${25}^{o}C$ are 200 mm Hg and 415 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of $CHC{l}_3$ and 40 g of $C{H}_{2}C{l}_2$ at the same temperature will be:

[Molecular mass of $CHC{l}_3=119.5g/mol$ and molecular mass of $C{H}_{2}C{l}_2=85g/mol$]

• A. 615.0 mm Hg
• B. 347.9 mm Hg
• C. 285.5 mm Hg
• D. 173.9 mm Hg

Answer 1

Answer: (B)

Molar mass of $C{H}_{2}C{l}_2$ $=12\times 1+1\times 2+35.5\times 2=85$ g mol${}^{–1}$

Molar mass of $CHC{l}_3=12\times 1+1\times 1+35.5\times 3=119.5$ g mol${}^{-1}$

Moles of $C{H}_{2}C{l}_2$= 40 g /85 g mol${}^{-1}$ = 0.47 mol

Moles of $CHC{l}_3$ = 25.5 g /119.5 g mol${}^{-1}$ = 0.213 mol

Total number of moles $=0.47+0.213=0.683$ mol

Mole fraction of component 2

$=0.47mol/0.683mol=0.688$

Mole fraction of component 1

$=1.00–0.688=0.312$

We know that:

$P_T=p_1^0+(p_2^0-p_1^0)x_2$

$=200+(415–200)\times 0.688$

$=200+147.9$

$=347.9$ mm Hg