The aqueous 0.01 Molal solution of `[Cr(NH_(3))_(6)]_(2)[Co(NH_(3))(NO_(2))_(5)]_(3)` is expected to have `DeltaT_(f)` equal to Given : `K_(f)` of `H_(2)O` is `"1.86 K kg mol"^(-1)`.

Options

A. 0.0186
B. 0.0372
C. 0.558
D. 0.093

Answer

To solve the problem, we need to calculate the depression in freezing point (ΔTf) of the given aqueous solution of the coordination compound \([Cr(NH_3)_6]_2[Co(NH_3)(NO_2)_5]_3\).

### Step-by-Step Solution:

1. **Identify the Formula for Depression in Freezing Point**:
The depression in freezing point 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 breaks into)
- \( K_f \) = freezing point depression constant of the solvent (for water, \( K_f = 1.86 \, \text{K kg mol}^{-1} \))
- \( m \) = molality of the solution

2. **Determine the Molality (m)**:
The problem states that the molality of the solution is \( 0.01 \, \text{molal} \).

3. **Calculate the van 't Hoff Factor (i)**:
The coordination compound \([Cr(NH_3)_6]_2[Co(NH_3)(NO_2)_5]_3\) dissociates into ions in solution.
- The dissociation can be represented as:
\[ [Cr(NH_3)_6]_2^{3+} + [Co(NH_3)(NO_2)_5]^{2-} \]
- This gives us:
- 2 ions from \([Cr(NH_3)_6]^{3+}\) (which dissociates into 2 Cr ions)
- 1 ion from \([Co(NH_3)(NO_2)_5]^{2-}\)
- Therefore, the total number of ions (n) after dissociation is:
\[ n = 2 + 3 = 5 \]
- Thus, the van 't Hoff factor \( i = 5 \).

4. **Substitute Values into the Formula**:
Now, we can substitute the values into the formula:
\[ \Delta T_f = i \cdot K_f \cdot m = 5 \cdot 1.86 \, \text{K kg mol}^{-1} \cdot 0.01 \, \text{molal} \]

5. **Calculate ΔTf**:
\[ \Delta T_f = 5 \cdot 1.86 \cdot 0.01 = 0.093 \, \text{K} \]

### Final Answer:
The expected depression in freezing point (\( \Delta T_f \)) is \( 0.093 \, \text{K} \).