A solution containing 0.10 g of non-volatile solute X (molar mass : 100) in 200 g of benzene depresses the freezing point of benzene by `0.25^(@)C` while 0.50 g of another non-volatile solute Y in 100 g of benzene also depresses the freezing point of benzene by `0.25^(@)C`. What is the molecular mass of Y ?

Options

A. 50
B. 100
C. 150
D. 1000

Explanation

To solve the problem, we will use the formula for freezing point depression, which is given by:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \(\Delta T_f\) is the depression in freezing point,
- \(K_f\) is the cryoscopic constant of the solvent (benzene in this case),
- \(m\) is the molality of the solution.

### Step 1: Calculate the molality of solute X

1. **Given data for solute X:**
- Mass of solute X (\(w_1\)) = 0.10 g
- Molar mass of solute X = 100 g/mol
- Mass of solvent (benzene) = 200 g = 0.200 kg

2. **Calculate the number of moles of solute X:**
\[ \text{Moles of X} = \frac{w_1}{\text{Molar mass of X}} = \frac{0.10 \, \text{g}}{100 \, \text{g/mol}} = 0.001 \, \text{mol} \]

3. **Calculate the molality of the solution:**
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.001 \, \text{mol}}{0.200 \, \text{kg}} = 0.005 \, \text{mol/kg} \]

### Step 2: Relate freezing point depression to molality for solute X

4. **Using the freezing point depression:**
\[ \Delta T_f = 0.25^\circ C \]
\[ 0.25 = K_f \cdot 0.005 \]
(We will need to find \(K_f\) later, but we can express it in terms of the molality.)

### Step 3: Calculate the molality of solute Y

5. **Given data for solute Y:**
- Mass of solute Y (\(w_2\)) = 0.50 g
- Mass of solvent (benzene) = 100 g = 0.100 kg
- The freezing point depression for Y is also \(0.25^\circ C\).

6. **Calculate the molality of the solution for Y:**
\[ \Delta T_f = K_f \cdot m_Y \]
\[ 0.25 = K_f \cdot m_Y \]

### Step 4: Relate the two equations

7. **Since both depressions are equal, we can set up the equation:**
\[ K_f \cdot 0.005 = K_f \cdot m_Y \]
This implies:
\[ m_Y = 0.005 \, \text{mol/kg} \]

### Step 5: Calculate the number of moles of solute Y

8. **Calculate the number of moles of solute Y:**
\[ \text{Moles of Y} = \frac{w_2}{\text{Molar mass of Y}} = \frac{0.50 \, \text{g}}{M_Y} \]

### Step 6: Set up the equation for molality of Y

9. **Using the definition of molality:**
\[ m_Y = \frac{\text{Moles of Y}}{\text{Mass of solvent in kg}} = \frac{0.50/M_Y}{0.100} \]
\[ 0.005 = \frac{0.50}{0.100 \cdot M_Y} \]
\[ 0.005 = \frac{5}{M_Y} \]

### Step 7: Solve for the molar mass of Y

10. **Rearranging gives:**
\[ M_Y = \frac{5}{0.005} = 1000 \, \text{g/mol} \]

### Conclusion

The molecular mass of solute Y is **1000 g/mol**.