10 gram of a non -volatile solute when dissolved in 100 gram of benzene raises its boiling point `1^(@)`. What is the molecular mass of the solute ? (`k_(b)` for benzene ` K mol^(-1)`).

Explanation

To find the molecular mass of the non-volatile solute, we can use the formula for elevation in boiling point:

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

Where:
- \(\Delta T_b\) is the elevation in boiling point (in °C or K),
- \(K_b\) is the ebullioscopic constant of the solvent (in K kg/mol),
- \(m\) is the molality of the solution (in moles of solute per kg of solvent).

### Step 1: Identify the given values
- Mass of the solute = 10 g
- Mass of the solvent (benzene) = 100 g = 0.1 kg (convert grams to kilograms)
- Elevation in boiling point (\(\Delta T_b\)) = 1 °C
- \(K_b\) for benzene = 2.53 K kg/mol

### Step 2: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. First, we need to calculate the number of moles of the solute.

To find the number of moles of solute, we use the formula:

\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molecular mass of solute (g/mol)}} \]

Let \(M\) be the molecular mass of the solute. Then:

\[ \text{Moles of solute} = \frac{10}{M} \]

Now, substituting this into the molality formula:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{10/M}{0.1} = \frac{100}{M} \]

### Step 3: Substitute into the boiling point elevation formula
Now we can substitute \(m\) into the elevation in boiling point formula:

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

Substituting the known values:

\[ 1 = 2.53 \cdot \frac{100}{M} \]

### Step 4: Solve for the molecular mass (M)
Rearranging the equation to solve for \(M\):

\[ M = 2.53 \cdot 100 \]

Calculating this gives:

\[ M = 253 \text{ g/mol} \]

### Conclusion
The molecular mass of the solute is **253 g/mol**.

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