Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?

Options

A. 20
B. 50
C. 100
D. 200

Explanation

To solve the problem, we will use the concept of relative lowering of vapor pressure and the relationships between the masses, molar masses, and mole fractions of the solute and solvent.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure liquid X (P₀) = 2 atm
- Vapor pressure of the solution (Pₛ) = 1 atm
- Mass of solute Y (W₂) = 1 g
- Mass of solvent X (W₁) = 20 g
- Molar mass of solvent X (M₁) = 200 g/mol

2. **Calculate Relative Lowering of Vapor Pressure:**
The relative lowering of vapor pressure is given by the formula:
\[ \frac{P₀ - Pₛ}{P₀} = X₂ \]
Substituting the values:
\[ \frac{2 \, \text{atm} - 1 \, \text{atm}}{2 \, \text{atm}} = X₂ \]
\[ \frac{1}{2} = X₂ \]

3. **Express Mole Fraction of Solute (X₂):**
The mole fraction of the solute can also be expressed as:
\[ X₂ = \frac{N₂}{N₁ + N₂} \]
Since the solution is dilute, we can approximate:
\[ X₂ \approx \frac{N₂}{N₁} \]
where \(N₂\) is the number of moles of solute Y and \(N₁\) is the number of moles of solvent X.

4. **Calculate Moles of Solvent (N₁):**
The number of moles of solvent X can be calculated using:
\[ N₁ = \frac{W₁}{M₁} = \frac{20 \, \text{g}}{200 \, \text{g/mol}} = 0.1 \, \text{mol} \]

5. **Calculate Moles of Solute (N₂):**
The number of moles of solute Y can be expressed as:
\[ N₂ = \frac{W₂}{M₂} = \frac{1 \, \text{g}}{M₂} \]

6. **Substitute into the Mole Fraction Equation:**
Now substituting \(N₁\) and \(N₂\) into the mole fraction equation:
\[ X₂ = \frac{N₂}{N₁} = \frac{\frac{1}{M₂}}{0.1} = \frac{10}{M₂} \]

7. **Set Up the Equation:**
We already found that \(X₂ = \frac{1}{2}\), so:
\[ \frac{10}{M₂} = \frac{1}{2} \]

8. **Solve for Molar Mass of Y (M₂):**
Cross-multiplying gives:
\[ 10 \cdot 2 = M₂ \implies M₂ = 20 \, \text{g/mol} \]

### Final Answer:
The molar mass of solute Y is **20 g/mol**.