Question

Vapour pressure of chloroform $\left( CHCl _{3}\right)$ and dichloromethane $\left( CH _{2} Cl _{2}\right)$ at $25^{\circ} C$ are $200\, mm$ of $Hg$ and $41.5\, mm$ of $Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5\, g$ of $CHCl _{3}$ and $40\, g$ of $CH _{2} Cl _{2}$ at the same temperature will be (Molecular mass of $CHCl _{3}=119.5\, u$ and molecular mass of $CH _{2} Cl _{2}=85\, u$ )

• 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 $CH _{2} Cl _{2}=12 \times 1+1 \times 2+35.5 \times 2=85 g mol ^{-1}$
Molar mass of $CHCl _{3}=12 \times 1+1 \times 1+35.5 \times 3=119.5 g mol ^{-1}$
Moles of $CH _{2} Cl _{2}=40 g / 85 g mol ^{-1}=0.47 mol$
Moles of $CHCl _{3}=25.5 g / 119.5 g mol ^{-1}=0.213 mol$
Total number of moles $=0.47+0.213=0.683 mol$
Mole fraction of component 2
$
=0.47 mol / 0.683 mol =0.688
$
Mole fraction of component 1
$
=1.00-0.688=0.312
$
We know that:
$
P _{ T }= p _{1}^{0}+\left( p _{2}^{0}- p _{1}^{0}\right) x _{2}
$
$
=200+(415-200) \times 0.688
$
$
=200+147.9
$
$
=347.9 mm Hg
$