JEE Main 2025 Jan 23 Shift 1, Chemistry Q4: Colligative Properties

JEE Main2025Jan 23, Shift 1Chemistry

$C r C l_{3} \cdot x N H_{3}$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $0.55 8^{\circ} C$ . Assuming 100% ionisation of this complex and coordination number of Cr is 6, the complex will be

(Given $K_{f} = 1.86$ K kg mol $^{- 1}$ )

A
$[C r (N H_{3})_{6}] C l_{3}$
B
$[C r (N H_{3})_{4} C l_{2}] C l$
C
$[C r (N H_{3})_{5} C l] C l_{2}$
D
$[C r (N H_{3})_{3} C l_{3}]$

Solution

Use the freezing-point depression equation $Δ T_{f} = i \cdot K_{f} \cdot m$ to extract the van't Hoff factor.

$0.558 = i \times 1.86 \times 0.1 \Rightarrow i = 3$

So the complex dissociates into 3 ions per formula unit. Checking the options:

$[C r (N H_{3})_{6}] C l_{3} \to 4$ ions, $[C r (N H_{3})_{4} C l_{2}] C l \to 2$ ions, $[C r (N H_{3})_{5} C l] C l_{2} \to 3$ ions, $[C r (N H_{3})_{3} C l_{3}] \to 1$ ion (neutral).

$∴$ The complex is $[C r (N H_{3})_{5} C l] C l_{2}$ .

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JEE Main 2025 Jan 23 Shift 1, Chemistry Q4: Colligative Properties

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