Ncert Exercises & Solutions

Using Raoult’s law explain how the total vapour pressure over the solution is related to mole fraction of components in the following solutions.

1. CHCl₃ (l) and CH₂Cl₂ (l)

2. NaCl (s) and H₂O (l)

According to Raoult's law for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

p_{1} = p_{1}^{0} x_{1}

(a) CHCl₃ (l) and CH₂Cl₂ (l) both are volatile components.

Hence, for a binary solution in which both components are volatile liquids, the total pressure will be

p = p₁ + p₂ = $x_{1} p_{1}^{0} + x_{2} p_{2}^{0}$

= $x_{1} p_{1}^{0}$ + (1 - x₁)

p_{2}^{0}

(p_{_{1}}^{0} - p_{2}^{0}) x_{1} + p_{2}^{0}

where, p = total vapour pressure
p₁ = partial vapour pressure of component1
p₂ = partial vapour pressure of component 2

(b) NaCl (s) and H₂O (l) both are non-volatile components.

Hence, for a solution containing non-volatile solute, the Raoult’s law is applicable only to vaporisable component and total vapour pressure can be written as

p = p₁= $x_{1} p_{1}^{0}$

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