Question

The vapour pressure of water at 20°C is 17.5 mm Hg. A solution of sucrose (molar
mass = 342) is prepared by dissolving 68.4 g in 1000 g of water. Calculate
(i) relative lowering of vapour pressure
(i) vapour pressure of solution.
​

Answer 1

By Raoult's Law, relative lowering of vapour pressure is :

$\dfrac{P^o_a-P_a}{P_a}=\dfrac{n_b}{n_a}=\dfrac{W_b\times M_a}{M_b\times W_a}$

Putting given values , we get :

$\dfrac{P^o_a-P_a}{P_a}=\dfrac{(68.4\ g)\times (18\ g/mol)}{(342\ g/mol)\times (1000\ g)}\\\\\dfrac{P^o_a-P_a}{P_a}=0.0036$  ......1)

Now, putting $P^o_a=17.5\ mm$ in equation 1 , we get :

$\dfrac{P^o_a}{P_a}-1=0.0036\\\\\dfrac{P^o_a}{P_a}=1.0036\\\\P_a=\dfrac{(17.5\ mm)}{1.00336}\\\\P_a=17.44\ mm$

Solution of each parts is (i) 0.0036 (ii) 17.44 mm.

Hence, this is the required solution.