A `0.01M` solution of glucose in water freezes at `-0.0186^(@)C`. A `0.01M` solution of KCI in water will freeze at temperature

Options

A. higher than `-0.0186^(@)C`
B. `0^(@)C`
C. `0.0186^(@)C`
D. lower than `-0.0186^(@)C`

Explanation

To solve the problem, we need to calculate the freezing point depression for a `0.01M` solution of KCl in water, using the information provided about the freezing point depression of a `0.01M` glucose solution.

### Step-by-Step Solution:

1. **Understand Freezing Point Depression**:
The freezing point depression is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) is the depression in freezing point,
- \(i\) is the van 't Hoff factor (number of particles the solute breaks into),
- \(K_f\) is the freezing point depression constant of the solvent (water in this case),
- \(m\) is the molality of the solution.

2. **Calculate \(K_f\) for Water**:
From the glucose solution, we know that:
- The freezing point of pure water is \(0°C\),
- The freezing point of the glucose solution is \(-0.0186°C\).

Therefore, the depression in freezing point for glucose is:
\[ \Delta T_f = 0 - (-0.0186) = 0.0186°C \]

For glucose, which is a non-electrolyte, \(i = 1\). Thus, we can rearrange the formula to find \(K_f\):
\[ 0.0186 = 1 \cdot K_f \cdot 0.01 \]
\[ K_f = \frac{0.0186}{0.01} = 1.86 \, °C \, kg/mol \]

3. **Determine the van 't Hoff Factor for KCl**:
KCl is an electrolyte and dissociates into two ions in solution:
\[ KCl \rightarrow K^+ + Cl^- \]
Therefore, the van 't Hoff factor \(i\) for KCl is \(2\).

4. **Calculate the Freezing Point Depression for KCl**:
Using the same formula for KCl:
\[ \Delta T_f = i \cdot K_f \cdot m \]
For KCl, \(m = 0.01\) (since it is also a \(0.01M\) solution), and substituting the values:
\[ \Delta T_f = 2 \cdot 1.86 \cdot 0.01 \]
\[ \Delta T_f = 0.0372°C \]

5. **Calculate the Freezing Point of KCl Solution**:
The freezing point of the KCl solution is:
\[ T_f = 0 - \Delta T_f = 0 - 0.0372 = -0.0372°C \]

### Final Answer:
The freezing point of a `0.01M` solution of KCl in water will be approximately \(-0.0372°C\).