Question

Vapour pressure of a pure liquid X is 2atm at 300K. it is lowered to 1atm on dissolving 1g of Y in 20g of liquid X. if the molar mass of X is 200. What is the molar mass of Y?
(A) 20
(B) 50
(C) 100
(D) 200

Answer 1

Hint: Generally, liquid solvents are volatile. The solute may be or may not be volatile. For solutions of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in the solution. This relationship is known as Raoult's law.

Complete step by step solution:
When a volatile solute is added to a volatile solvent, then vapour pressure of the solution decreases. There are many properties of solutions based on the decreasing the vapour pressure which is known as colligative properties.
These are, (1) relative lowering of the vapour pressure of the solvent
(2) Depression of freezing point of the solvent
(3) Elevation of a boiling point of the solvent
(4) Osmotic pressure.
The problem given is related to the property relative lowering of vapour pressure.
According to the property, the relative lowering of the lower pressure of the solvent is equal to the mole fraction of the solute.
$\dfrac{{{p}^{o}}-p}{{{p}^{o}}}=\dfrac{{{w}_{2}}X{{M}_{1}}}{{{M}_{2}}X{{w}_{1}}}$ ----- (1)
Here ${{w}_{1}}\And {{w}_{2}}$ are the mass and ${{M}_{1}}\And {{M}_{2}}$ are the molar mass of the solvent and solute respectively.
Given vapour pressure of pure liquid X, ${{p}^{o}}$ = 2 atm
Vapour pressure of the solution, p = 1 atm
Weight of solute, ${{w}_{2}}$ = 1g
Weight of solvent, ${{w}_{1}}$ = 20 g of pure liquid X
Molar mass of solvent pure liquid X, ${{M}_{1}}$ = 20 g
Molar mass of solute Y, ${{M}_{2}}=?$
Substitute the above values in the equation (1),
$\dfrac{2-1}{2}=\dfrac{1X200}{{{M}_{2}}X20}$
Therefore, ${{M}_{2}}=20g$
Hence, the molar mass of solute Y = 20g

The correct answer is option A.

Note: All of these colligative properties the number of solute particles irrespective of their nature related to the total number of particles present in the solution. These properties are used to determine the molar mass of solutes.