Question

The moles of $ Ba{(N{O_3})_2} $ which on dissolving in 12 moles of water lowers the vapour pressure of water to 80% are?

Answer 1

Hint: Roult’s law states that the vapour pressure of solvent in a solution at a given temperature will be directly proportional to the mole fraction of the solvent. The relative lowering of vapour pressure is a colligative property as its properties depend upon the mole fractions of the solute in a solution.

Complete answer:
When we add a non-volatile solute in a solution, the tendency of the solvent to escape into the vapour phase reduces. They reduce the vapour pressure of the solvent molecules. The relative lowering of vapour pressure is the ratio of lowering of vapour pressure to that of the vapour pressure of a pure solvent.
The expression for relative lowering of vapour pressure if given as: $ \dfrac{{P_1^0 - {P_1}}}{{P_1^0}} = {\chi _2} $
Where, $ P_1^0 - {P_1} $ = Lowering of vapour pressure, $ P_1^0 = $ vapour pressure of the pure solvent, and $ {\chi _2} = $ mole fraction of the solute.
The mole fraction of the solute is given by the formula: $ \dfrac{{{n_2}}}{{{n_1} + {n_2}}} $ (where $ {n_2} = $ no. of moles of solute and $ {n_1} = $ no. of moles of solvent.
The information given to us is: No. of moles of solvent = 12mol, $ P_1^0 = 100,{P_1} = 80 $ . Substituting in the formula;
 $ \dfrac{{100 - 80}}{{100}} = \dfrac{n}{{n + 12}} $
 $ \dfrac{{20}}{{100}} = \dfrac{n}{{n + 12}} $
 $ 100n = 20n + 240 $
 $ 80n = 240 \to n = \dfrac{{240}}{{80}} = 3 $
Therefore, the number of moles of solute is 3mol.

Note:
If the solution is very dilute $ {n_1} + {n_2} \approx {n_1} $ (since the moles of solvent are very large, we can ignore the no. of moles of solvent). The molar mass of the solute can be efficiently found using this equation as $ n = \dfrac{w}{M} $ (where w= mass of solute in g and M is the molar mass of the solute in g/mol). Therefore for dilute solutions the equation can be written as: $ \dfrac{{P_1^0 - {P_1}}}{{P_1^0}} = {n_2} = \dfrac{w}{M} $ .