Depression in freezing point for 1 M urea, 1 M NaCl and 1 M `CaCl_(2)` are in the ratio of

Options

A. `1:2:3`
B. `1:1:1`
C. `3:2:1`
D. Data insufficient

Answer

To solve the problem of finding the depression in freezing point for 1 M urea, 1 M NaCl, and 1 M CaCl₂ in the ratio, we will follow these steps:

### Step-by-Step Solution:

1. **Understand Depression in Freezing Point**:
Depression in freezing point (ΔTf) is a colligative property that depends on the number of solute particles in a solution. It can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = Van't Hoff factor (number of particles the solute dissociates into)
- \( K_f \) = freezing point depression constant (depends on the solvent)
- \( m \) = molality of the solution

2. **Identify the Van't Hoff Factors**:
- **For Urea**: Urea is a non-electrolyte and does not dissociate in solution. Thus, \( i = 1 \).
- **For NaCl**: Sodium chloride dissociates into two ions: Na⁺ and Cl⁻. Therefore, \( i = 2 \).
- **For CaCl₂**: Calcium chloride dissociates into three ions: Ca²⁺ and 2 Cl⁻. Thus, \( i = 3 \).

3. **Calculate the Ratios**:
Since the molality (m) and the freezing point depression constant (Kf) are the same for all three solutions, we can focus on the Van't Hoff factors to find the ratio of the depression in freezing points:
- For Urea: \( \Delta T_f = 1 \cdot K_f \cdot 1 = K_f \)
- For NaCl: \( \Delta T_f = 2 \cdot K_f \cdot 1 = 2K_f \)
- For CaCl₂: \( \Delta T_f = 3 \cdot K_f \cdot 1 = 3K_f \)

4. **Establish the Ratio**:
The ratios of the depression in freezing points for urea, NaCl, and CaCl₂ are:
\[ \Delta T_f \text{ (urea)} : \Delta T_f \text{ (NaCl)} : \Delta T_f \text{ (CaCl₂)} = K_f : 2K_f : 3K_f = 1 : 2 : 3 \]

5. **Conclusion**:
Therefore, the depression in freezing point for 1 M urea, 1 M NaCl, and 1 M CaCl₂ are in the ratio of:
\[ \text{Answer: } 1 : 2 : 3 \]