A solution containing 0.52 g of `C_(10)H_(8)` in 50g `CCl_(4)` produced an elevation in boiling point of `0.402^(@)C`. On the other hand a solution of 0.62 g of an unknown solute dissolved in same amount of `CCl_(4)` produced an elevation of `0.65^(@)C`. Molecular mass of solute is :

Options

A. 85.53 g/mole
B. 181.51 g/mole
C. 94.38 g/mole
D. 160.62 g/mole

Answer

To solve the problem, we need to find the molecular mass of the unknown solute based on the given data about the elevation in boiling point of two different solutions. Let's break down the steps:

### Step 1: Calculate the molality of naphthalene (C10H8) in CCl4
1. **Given data**:
- Mass of naphthalene (C10H8) = 0.52 g
- Molar mass of naphthalene (C10H8) = 128 g/mol
- Mass of CCl4 (solvent) = 50 g = 0.050 kg

2. **Calculate moles of naphthalene**:
\[ \text{Moles of naphthalene} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.52 \text{ g}}{128 \text{ g/mol}} = 0.0040625 \text{ mol} \]

3. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0040625 \text{ mol}}{0.050 \text{ kg}} = 0.08125 \text{ mol/kg} \]

### Step 2: Use the elevation in boiling point to find the ebullioscopic constant (K_b)
1. **Given elevation in boiling point (ΔT_b)**:
- ΔT_b for naphthalene = 0.402 °C

2. **Using the formula**:
\[ \Delta T_b = K_b \cdot m \]
Rearranging gives:
\[ K_b = \frac{\Delta T_b}{m} = \frac{0.402 \text{ °C}}{0.08125 \text{ mol/kg}} \approx 4.94 \text{ °C kg/mol} \]

### Step 3: Calculate the molality of the unknown solute
1. **Given data for unknown solute**:
- Mass of unknown solute = 0.62 g
- Elevation in boiling point (ΔT_b) = 0.65 °C

2. **Let the molar mass of the unknown solute be M**. The molality (m) is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{\frac{0.62 \text{ g}}{M}}{0.050 \text{ kg}} = \frac{0.62}{M \cdot 0.050} \]

### Step 4: Use the elevation in boiling point to find the molar mass of the unknown solute
1. **Using the formula again**:
\[ \Delta T_b = K_b \cdot m \]
Substituting the values:
\[ 0.65 = 4.94 \cdot \frac{0.62}{M \cdot 0.050} \]

2. **Rearranging to find M**:
\[ M = \frac{4.94 \cdot 0.62}{0.65 \cdot 0.050} \]
\[ M = \frac{3.0688}{0.0325} \approx 94.38 \text{ g/mol} \]

### Conclusion
The molecular mass of the unknown solute is approximately **94.38 g/mol**.