A solution containing 0.52 g of `C_(10)H_(8)` in `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 will use the formula for boiling point elevation, which is given by:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \(\Delta T_b\) = elevation in boiling point
- \(K_b\) = ebullioscopic constant of the solvent (which we will assume to be constant for both cases since the solvent is the same)
- \(m\) = molality of the solution

### Step 1: Calculate the number of moles of \(C_{10}H_8\)

Given:
- Mass of \(C_{10}H_8\) = 0.52 g
- Molar mass of \(C_{10}H_8\) = 128 g/mol

The number of moles (\(n\)) can be calculated using the formula:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.52 \, \text{g}}{128 \, \text{g/mol}} = 0.0040625 \, \text{mol} \]

### Step 2: Set up the boiling point elevation equation for \(C_{10}H_8\)

Using the boiling point elevation formula:

\[ \Delta T_b = K_b \cdot m \]

Where \(m\) (molality) is defined as:

\[ m = \frac{n}{\text{mass of solvent in kg}} \]

Let the mass of solvent (CCl4) be \(m_{CCl4}\) kg. Then:

\[ 0.402 = K_b \cdot \frac{0.0040625}{m_{CCl4}} \]

Rearranging gives:

\[ K_b = \frac{0.402 \cdot m_{CCl4}}{0.0040625} \]

### Step 3: Set up the boiling point elevation equation for the unknown solute

For the unknown solute, we have:

Given:
- Mass of unknown solute = 0.62 g
- Elevation in boiling point = 0.65°C

Using the same boiling point elevation formula:

\[ 0.65 = K_b \cdot \frac{0.62/x}{m_{CCl4}} \]

Where \(x\) is the molar mass of the unknown solute.

Rearranging gives:

\[ K_b = \frac{0.65 \cdot m_{CCl4} \cdot x}{0.62} \]

### Step 4: Equate the two expressions for \(K_b\)

Now we have two expressions for \(K_b\):

1. \(K_b = \frac{0.402 \cdot m_{CCl4}}{0.0040625}\)
2. \(K_b = \frac{0.65 \cdot m_{CCl4} \cdot x}{0.62}\)

Setting them equal to each other:

\[ \frac{0.402 \cdot m_{CCl4}}{0.0040625} = \frac{0.65 \cdot m_{CCl4} \cdot x}{0.62} \]

### Step 5: Cancel \(m_{CCl4}\) and solve for \(x\)

Since \(m_{CCl4}\) appears on both sides, we can cancel it out (assuming it's not zero):

\[ \frac{0.402}{0.0040625} = \frac{0.65 \cdot x}{0.62} \]

Cross-multiplying gives:

\[ 0.402 \cdot 0.62 = 0.65 \cdot x \cdot 0.0040625 \]

Calculating the left side:

\[ 0.402 \cdot 0.62 = 0.24924 \]

And solving for \(x\):

\[ x = \frac{0.24924}{0.65 \cdot 0.0040625} \]

Calculating the right side:

\[ x = \frac{0.24924}{0.002640625} \approx 94.38 \, \text{g/mol} \]

### Final Answer

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

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