The boiling point of water `(100^(@)C)` becomes `100.52^(@)` C if 3 g of a non - volatile solute is dissolved in 20 ml of it . Calculate the molar mass of the solute (`k_(b)` for water ` = 0.52 K m^(-1)`).

Explanation

To solve the problem, we need to follow a series of steps to calculate the molar mass of the non-volatile solute based on the given information about the boiling point elevation.

### Step-by-Step Solution:

1. **Identify the Change in Boiling Point (ΔTb)**:
The boiling point of water increases from \(100^\circ C\) to \(100.52^\circ C\). Therefore, the change in boiling point is:
\[ \Delta T_b = 100.52^\circ C - 100^\circ C = 0.52^\circ C \]

2. **Use the Boiling Point Elevation Formula**:
The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \(i\) = van 't Hoff factor (which is 1 for non-volatile solutes that do not dissociate)
- \(K_b\) = ebullioscopic constant for water = \(0.52 \, K \cdot m^{-1}\)
- \(m\) = molality of the solution

3. **Substitute Known Values**:
Since the solute is non-volatile, we can assume \(i = 1\). Thus, we can rewrite the equation as:
\[ 0.52 = 1 \cdot 0.52 \cdot m \]
Simplifying this gives:
\[ m = 1 \, mol/kg \]

4. **Convert the Volume of Water to Mass**:
We are given that 20 mL of water is used. The density of water is approximately \(1 \, g/mL\), so:
\[ \text{Mass of water} = 20 \, g \]

5. **Convert Mass of Water to Kilograms**:
To convert grams to kilograms:
\[ \text{Mass of water in kg} = \frac{20 \, g}{1000} = 0.02 \, kg \]

6. **Calculate the Number of Moles of Solute**:
The molality \(m\) is defined as:
\[ m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} \]
Rearranging the formula gives:
\[ \text{Number of moles of solute} = m \cdot \text{Mass of solvent in kg} = 1 \cdot 0.02 = 0.02 \, mol \]

7. **Calculate the Molar Mass of the Solute**:
The number of moles of solute can also be expressed as:
\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
We know the mass of the solute is 3 g, so:
\[ 0.02 = \frac{3}{M} \]
Rearranging gives:
\[ M = \frac{3}{0.02} = 150 \, g/mol \]

### Final Answer:
The molar mass of the solute is \(150 \, g/mol\).