When 50 g of a nonvolatile solute is dissolved in a certain quantity of solvent, the elevation of boiling point is 2.0 K. What will be the elevation of boiling point when 30 g of solute is dissolved in the same amount of the same solvent?

Answer

To solve the problem of boiling point elevation when different amounts of a non-volatile solute are dissolved in the same solvent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula**: The elevation of boiling point (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \cdot m \]
where \( K_b \) is the ebullioscopic constant of the solvent and \( m \) is the molality of the solution.

2. **Calculate Molality for 50 g of Solute**:
Let's denote the molar mass of the solute as \( M \). The molality (\( m \)) can be calculated as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Given that 50 g of solute is dissolved, the number of moles of solute is:
\[ \text{moles} = \frac{50 \, \text{g}}{M} \]
Assuming we have 1 kg of solvent, the molality when 50 g of solute is used becomes:
\[ m_{50} = \frac{50/M}{1} = \frac{50}{M} \]

3. **Calculate the Elevation of Boiling Point for 50 g**:
Using the boiling point elevation formula:
\[ \Delta T_b = K_b \cdot \frac{50}{M} \]
We know that \( \Delta T_b \) for 50 g is 2.0 K, so we can write:
\[ 2.0 = K_b \cdot \frac{50}{M} \quad \text{(1)} \]

4. **Calculate Molality for 30 g of Solute**:
Now, for 30 g of solute, the number of moles is:
\[ \text{moles} = \frac{30 \, \text{g}}{M} \]
The molality when 30 g of solute is used is:
\[ m_{30} = \frac{30/M}{1} = \frac{30}{M} \]

5. **Calculate the Elevation of Boiling Point for 30 g**:
Using the same boiling point elevation formula:
\[ \Delta T_b = K_b \cdot \frac{30}{M} \quad \text{(2)} \]

6. **Relate the Two Elevations**:
We can relate the two equations (1) and (2):
\[ \frac{\Delta T_b \text{ (for 30 g)}}{\Delta T_b \text{ (for 50 g)}} = \frac{m_{30}}{m_{50}} \]
Substituting the values:
\[ \frac{\Delta T_b \text{ (for 30 g)}}{2.0} = \frac{30/M}{50/M} \]
Simplifying gives:
\[ \frac{\Delta T_b \text{ (for 30 g)}}{2.0} = \frac{30}{50} = \frac{3}{5} \]

7. **Calculate \( \Delta T_b \) for 30 g**:
Rearranging gives:
\[ \Delta T_b \text{ (for 30 g)} = 2.0 \cdot \frac{3}{5} = 1.2 \, \text{K} \]

### Final Answer:
The elevation of boiling point when 30 g of solute is dissolved in the same amount of solvent is **1.2 K**.