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? `(K_f for H_2O = 1.86°Cm^-1)` and K_b 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 determine the boiling point of a dilute solution of a non-volatile solute in water, given that it freezes at -0.2°C. We will use the freezing point depression and boiling point elevation formulas.

### Step-by-Step Solution:

1. **Identify the given values:**
- Freezing point depression (ΔTf) = 0.2°C (since the normal freezing point of water is 0°C, and the solution freezes at -0.2°C).
- Kf (freezing point depression constant for water) = 1.86°C/m.
- Kb (boiling point elevation constant for water) = 0.515°C/m.

2. **Use the relationship between ΔTf and ΔTb:**
The relationship between freezing point depression and boiling point elevation is given by:
\[ \frac{\Delta Tf}{\Delta Tb} = \frac{Kf}{Kb} \]
Rearranging this gives us:
\[ \Delta Tb = \Delta Tf \times \frac{Kb}{Kf} \]

3. **Substitute the known values into the equation:**
\[ \Delta Tb = 0.2 \times \frac{0.515}{1.86} \]

4. **Calculate ΔTb:**
First, calculate the fraction:
\[ \frac{0.515}{1.86} \approx 0.276 \]
Now, multiply by 0.2:
\[ \Delta Tb \approx 0.2 \times 0.276 \approx 0.0552°C \]

5. **Determine the boiling point of the solution:**
The normal boiling point of water is 100°C. Therefore, the boiling point of the solution (Tb) can be calculated as:
\[ Tb = 100°C + \Delta Tb \]
\[ Tb \approx 100°C + 0.0552°C \approx 100.0552°C \]

6. **Final Answer:**
The boiling point of the solution is approximately **100.06°C** (rounded to two decimal places).