The boiling point of benzene is 353 .23 K. When 1.80 gram of non- volatile solute was dissolved in 90 gram of benzene , the boiling point is raised to 354.11 K. Calculate the molar mass of solute .
`[K_(b)`for benzene = 2.53 K kg `mol^(-1)]`

Explanation

To calculate the molar mass of the non-volatile solute, we will use the formula for boiling point elevation:

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

Where:
- \(\Delta T_b\) = change in boiling point
- \(K_b\) = ebullioscopic constant of the solvent (benzene in this case)
- \(m\) = molality of the solution

### Step 1: Calculate the change in boiling point (\(\Delta T_b\))

Given:
- Boiling point of pure benzene (\(T_{b, \text{pure}}\)) = 353.23 K
- Boiling point of the solution (\(T_{b, \text{solution}}\)) = 354.11 K

\[ \Delta T_b = T_{b, \text{solution}} - T_{b, \text{pure}} = 354.11 \, \text{K} - 353.23 \, \text{K} = 0.88 \, \text{K} \]

### Step 2: Calculate the molality (m)

We know that:

\[ m = \frac{W_2}{W_1} \cdot \frac{1000}{M_2} \]

Where:
- \(W_2\) = mass of the solute = 1.80 g
- \(W_1\) = mass of the solvent (benzene) = 90 g
- \(M_2\) = molar mass of the solute (unknown)

Rearranging the molality formula gives:

\[ m = \frac{1.80 \, \text{g}}{90 \, \text{g}} \cdot \frac{1000}{M_2} \]

### Step 3: Substitute into the boiling point elevation formula

Substituting the values into the boiling point elevation formula:

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

Substituting for \(m\):

\[ 0.88 = 2.53 \cdot \left(\frac{1.80}{90} \cdot \frac{1000}{M_2}\right) \]

### Step 4: Solve for \(M_2\)

Rearranging the equation to solve for \(M_2\):

\[ 0.88 = 2.53 \cdot \left(\frac{1.80 \cdot 1000}{90 \cdot M_2}\right) \]

\[ M_2 = \frac{2.53 \cdot 1.80 \cdot 1000}{0.88 \cdot 90} \]

Calculating the right-hand side:

\[ M_2 = \frac{2.53 \cdot 1.80 \cdot 1000}{79.2} \]

\[ M_2 = \frac{4554}{79.2} \approx 57.5 \, \text{g/mol} \]

### Final Result

The molar mass of the non-volatile solute is approximately **57.5 g/mol**.

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