The vapour pressure of pure ether (molar mass 74 g/mol) at 293k is 588 m bar. A 10% (w/w) solutions of benzoic acid (molar mass=122 g/mol) was prepared in ether at 1293k. What will be the vapour pressure of solutions?

The total Vapour pressure will be:

$P_T = {p_1}^0{x}_1 + {p_2}^0{x}_2$

Where P₁⁰ = V.P of pure ether
X₁ = mole fraction of ether

P₂⁰ = V.P of benzoic acid
X₂ = mole fraction of benzoic acid

However as benzioc acid is non-volatile the value of P₂⁰ will be 0

Hence lowering of V.P will be:

$$\begin{gathered} ∆P = {P_1}^0 (1-x_1) \\ = {P_1}^0 x_2 \end{gathered}$$

So
X₂ = moles of benzioc acid/ (moles of ether + moles of Benzoic acid)

Assuming we have 100 g of mixture.

So wt. of Benzoic acid = 10 gm and remaing 90 g is ether

No. of moles of benzioc acid = 10/122 = 0.081
No of moles of ether = 90/74 = 1.216

Total no of moles = 0.081 + 1.216 = 1.297

Mole fraction of Benzioc accid = 0.081/ 1.297 = 0.0624

Thus the decrease in V.P = 588 x 0.0624
= 36.69 m bar

Thus the final V.P = 588 - 36.69 =  551.31 m Bar