Verified Solution • 2025–26 Session moderate

QJEEM25Chb25c: JEE Main 2025 Chemistry - Class 12 JEEM25Chb25c Solution: can exist as a complex. 0.1 molal aqueous solution of this…

CrCl3⋅xNH3\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∘C0.558^{\circ} \mathrm{C} . Assuming 100%100 \% ionisation of this complex and coordination number of Cr is 6 , the complex will be (Given Kf=1.86 K kg mol−1\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} )

Multiple Choice Options

  • A [Cr(NH3)5Cl]Cl2\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2
  • B [Cr(NH3)4Cl2]Cl\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}
  • C [Cr(NH3)6]Cl3\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3
  • D [Cr(NH3)3Cl3]\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_3 \mathrm{Cl}_3\right]

Step-by-Step Verified Solution

amp; Given : ΔTf=0.558∘Camp;amp;kf=1.86 K×kg molamp;0.1 m aq. sol.⇒amp;ΔTf=i×kf×m⇒amp;0.558=i×1.86×0.1⇒amp;i=3\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}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring