A dilute solution of a non-volatile solute in water freezes at — 0.2°C. At what temperature (in °C) the same solution will boil? (Kt for `H_2O` = 1.86°C m`^(-1)`and K0 for `H_2O` = 0 515°C `m^(-1)`)

Options

A. 100.121
B. 100.217
C. 100.056
D. 101.562

Answer

To solve the problem, we need to find the boiling point of a dilute solution of a non-volatile solute in water, given that the freezing point of the solution is -0.2°C. We will use the freezing point depression and boiling point elevation formulas.

### Step-by-Step Solution:

1. **Identify the Freezing Point Depression**:
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{solvent}} - T_f^{\text{solution}} \]
Here, \( T_f^{\text{solvent}} \) (the freezing point of pure water) is 0°C, and \( T_f^{\text{solution}} \) is -0.2°C.
\[ \Delta T_f = 0°C - (-0.2°C) = 0.2°C \]

**Hint**: Remember that the freezing point of the solvent is always higher than that of the solution when a solute is added.

2. **Calculate the Molality (m)**:
We can use the freezing point depression formula:
\[ \Delta T_f = K_f \cdot m \]
Where \( K_f \) for water is given as 1.86°C/m. Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} = \frac{0.2°C}{1.86°C/m} \approx 0.1075 \, m \]

**Hint**: Make sure to use the correct units for \( K_f \) and \( \Delta T_f \) when calculating molality.

3. **Calculate the Boiling Point Elevation**:
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:
\[ \Delta T_b = K_b \cdot m \]
Where \( K_b \) for water is given as 0.515°C/m. Substituting the values:
\[ \Delta T_b = 0.515°C/m \cdot 0.1075 \, m \approx 0.05537°C \]

**Hint**: The boiling point elevation is directly proportional to the molality of the solution.

4. **Calculate the New Boiling Point of the Solution**:
The boiling point of the solution can be found by adding the boiling point elevation to the boiling point of pure water (100°C):
\[ T_b^{\text{solution}} = T_b^{\text{solvent}} + \Delta T_b = 100°C + 0.05537°C \approx 100.05537°C \]

**Hint**: Always remember to add the boiling point elevation to the boiling point of the pure solvent to find the boiling point of the solution.

5. **Final Answer**:
Rounding the result, the boiling point of the solution is approximately:
\[ T_b^{\text{solution}} \approx 100.056°C \]

Thus, the answer is **100.056°C**.