The freezing point of water is depressed by `0.37^(@)C` in a 0.01 molar NaCI solution. The freezing point of 0.02 molar solution is depressed by

Options

A. `0.37^(@)C`
B. `0.74^(@)C`
C. `0.185^(@)C`
D. `0^(@)C`

Explanation

To solve the problem, we need to calculate the depression in freezing point for a 0.02 molar NaCl solution based on the information given for a 0.01 molar NaCl solution.

### Step-by-Step Solution:

1. **Understanding Freezing Point Depression**:
The freezing point depression (\( \Delta T_f \)) is a colligative property, which means it depends on the number of solute particles in a solution, not on the identity of the solute. The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = cryoscopic constant of the solvent (for water, \( K_f \) is approximately 1.86 °C kg/mol)
- \( m \) = molality of the solution

2. **Given Data**:
- For a 0.01 molar NaCl solution, the freezing point depression is \( \Delta T_f = 0.37 \) °C.
- The molarity of the solution is 0.01 mol/L.

3. **Calculate \( i \cdot K_f \)**:
Since NaCl dissociates into 2 ions (Na\(^+\) and Cl\(^-\)), the van 't Hoff factor \( i \) for NaCl is 2.
We can rearrange the formula to find \( i \cdot K_f \):
\[ 0.37 = i \cdot K_f \cdot 0.01 \]
\[ i \cdot K_f = \frac{0.37}{0.01} = 37 \]

4. **Calculate Depression for 0.02 M NaCl Solution**:
Now, we need to find the depression in freezing point for a 0.02 molar NaCl solution.
Using the same formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Here, \( m = 0.02 \) mol/L.
Substituting the values:
\[ \Delta T_f = 37 \cdot 0.02 = 0.74 \text{ °C} \]

5. **Conclusion**:
The freezing point of the 0.02 molar NaCl solution is depressed by \( 0.74 \) °C.

### Final Answer:
The freezing point of a 0.02 molar NaCl solution is depressed by **0.74 °C**.