If 30 g a solute of molecular mass 154 is dissolved in 250 g of benzene. What will be the elevation in boiling point of the resuling solution ?
(Given : `K_B(C_(6)H_(6)) =2.6 K kg mol^(-1)`)

Answer

To find the elevation in boiling point of the solution, we can use the formula:

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

where:
- \(\Delta T_b\) is the elevation in boiling point,
- \(K_b\) is the ebullioscopic constant of the solvent (benzene in this case),
- \(m\) is the molality of the solution.

### Step 1: Calculate the number of moles of solute
The number of moles of solute can be calculated using the formula:

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

Given:
- Mass of solute = 30 g
- Molecular mass of solute = 154 g/mol

\[ \text{Number of moles} = \frac{30 \, \text{g}}{154 \, \text{g/mol}} \approx 0.1942 \, \text{mol} \]

### Step 2: Calculate the mass of the solvent in kilograms
The mass of the solvent (benzene) is given as 250 g. We need to convert this to kilograms:

\[ \text{Mass of solvent} = 250 \, \text{g} = \frac{250}{1000} \, \text{kg} = 0.25 \, \text{kg} \]

### Step 3: Calculate the molality of the solution
Molality \(m\) is defined as the number of moles of solute per kilogram of solvent:

\[ m = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} \]

Substituting the values we calculated:

\[ m = \frac{0.1942 \, \text{mol}}{0.25 \, \text{kg}} \approx 0.7768 \, \text{mol/kg} \]

### Step 4: Calculate the elevation in boiling point
Now we can substitute the values of \(K_b\) and \(m\) into the elevation in boiling point formula:

Given:
- \(K_b\) for benzene = 2.6 K kg/mol

\[ \Delta T_b = K_b \times m = 2.6 \, \text{K kg/mol} \times 0.7768 \, \text{mol/kg} \approx 2.02 \, \text{K} \]

### Final Answer
The elevation in boiling point of the resulting solution is approximately:

\[ \Delta T_b \approx 2.02 \, \text{K} \]

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