# Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2CI2) at 25oC are 200 mmHg and 41.5 mmHg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCI3 and 40 g of CH2CI2 at the same temperature will be :(Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2CI2 = 85 u) Option 1) 615.0 mmHg Option 2) 347.9 mmHg Option 3) 285.5 mmHg Option 4) 173.9 mmHg

As we learnt in

Rault's Law -

The total vapour pressure of binary mixture of miscible liquids be having ideally is given by

$P_{T}= P_{A}^{0}x_{A}+P_{B}^{0}x_{B}$

Where $x_{A}$  and  $x_{B}$ are mole fraction of A and B in liquid phase.

- wherein

$P_{A}^{0}$  and $P_{B}^{0}$ are vapour pressures of pure liquids.

Moles of $CHCl_{3}=\frac{25.5}{119.5}=0.213$

Moles of $CH_{2}Cl_{2}=\frac{40}{85}=0.471$

Mole fraction $X_{CHCl_{3}}=\frac{0.213}{0.471+0.213}=\frac{0.213}{0.684}=0.311$

$X_{CH_{2}Cl_{2}}=\frac{0.471}{0.684}=0.689$

Vapour pressure = $0.311\times P_{CHCl_{3}}+0.689\ P_{CH_{2}Cl_{2}}$

$=0.311\times 200 + 0.689 \times 41.5=90.7935\ mm Hg$

The closest option is 173.9 mm Hg.

Option 1)

615.0 mmHg

Incorrect

Option 2)

347.9 mmHg

Incorrect

Option 3)

285.5 mmHg

Incorrect

Option 4)

173.9 mmHg

Correct

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