CrCl 3 × xNH 3 can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558 ° 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 )

Chemistry · Ionic Equilibrium

JEE Main 2025 — 23 January, Morning Shift — Question 29

$\mathrm{CrCl}_{3} \cdot \mathrm{xNH}_{3}$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $0.558^{\circ} \mathrm{C}$ . Assuming $100 \%$ ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Option A:
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{6}\right] \mathrm{Cl}_{3}$
Option B:
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}$
Option C:
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}\right] \mathrm{Cl}_{2}$
Correct
Option D:
$\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{3} \mathrm{Cl}_{3}\right]$

Answer: C

Step-by-step solution

$\Delta \mathrm{T}_{\mathrm{f}}=0.558^{\circ} \mathrm{C}$

\mathrm{k}_{\mathrm{f}}=1.86 \frac{\mathrm{~K} \times \mathrm{kg}}{\mathrm{~mol}}

0.1 m aq. sol.

$\Rightarrow \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{k}_{\mathrm{f}} \times \mathrm{m}$

$\Rightarrow 0.558=\mathrm{i} \times 1.86 \times 0.1$

$\Rightarrow \mathrm{i}=3$

Answer key and solution verified before publishing.

Practise Ionic Equilibrium

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 23 January, Morning Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: Sparingly Soluble Salts, Solubility Product & Precipitation Conditions

Answer: C

Step-by-step solution

Practise Ionic Equilibrium

More Ionic Equilibrium questions from this paper