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 freezing point depression (\( \Delta T_f \)) of the given complex compound in an aqueous solution. The formula to calculate the freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \( \Delta T_f \) = freezing point depression
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_f \) = cryoscopic constant of the solvent (for water, \( K_f = 1.86 \, \text{K kg mol}^{-1} \))
- \( m \) = molality of the solution

### Step 1: Identify the molality and \( K_f \)
The molality (\( m \)) of the solution is given as \( 0.01 \, \text{molal} \), and the \( K_f \) for water is \( 1.86 \, \text{K kg mol}^{-1} \).

### Step 2: Determine the dissociation of the complex
The complex given is \([Cr(NH_3)_6]_2[Co(NH_3)(NO_2)_5]_3\).

When this complex dissociates in solution, it breaks down into:
- 2 ions of \([Cr(NH_3)_6]^{3+}\)
- 3 ions of \([Co(NH_3)(NO_2)_5]^{2-}\)

Thus, the total number of ions produced is:
- 2 (from \([Cr(NH_3)_6]^{3+}\)) + 3 (from \([Co(NH_3)(NO_2)_5]^{2-}\)) = 5 ions.

### Step 3: Calculate the van 't Hoff factor \( i \)
The van 't Hoff factor \( i \) is the total number of particles in solution after dissociation. Here, \( i = 5 \).

### Step 4: Substitute values into the freezing point depression formula
Now we can substitute the values into the formula:

\[ \Delta T_f = i \cdot K_f \cdot m \]
\[ \Delta T_f = 5 \cdot 1.86 \cdot 0.01 \]

### Step 5: Perform the calculation
Calculating the above expression:

\[ \Delta T_f = 5 \cdot 1.86 \cdot 0.01 = 0.093 \]

### Conclusion
Thus, the expected freezing point depression \( \Delta T_f \) for the solution is \( 0.093 \, \text{K} \).

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