JEE MAIN - Chemistry (2025 - 23rd January Morning Shift - No. 6)
$\mathrm{CrCl}_3 \cdot \mathrm{xNH}_3$ 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 $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2$
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$
$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]$
Explanation
$$\begin{aligned}
&\text { Given : } \Delta \mathrm{T}_{\mathrm{f}}=0.558^{\circ} \mathrm{C}\\
&\begin{aligned}
& \mathrm{k}_{\mathrm{f}}=1.86 \frac{\mathrm{~K} \times \mathrm{kg}}{\mathrm{~mol}} \\
& 0.1 \mathrm{~m} \text { aq. } \mathrm{sol} . \\
\Rightarrow & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
\Rightarrow & 0.558=\mathrm{i} \times 1.86 \times 0.1 \\
\Rightarrow & \mathrm{i}=3
\end{aligned}
\end{aligned}$$
Comments (0)
