MCQ
$CrCl _3 \cdot xNH _3$ can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of $0.558^{\circ} C$. 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
    $\left[ Cr \left( NH _3\right)_6\right] Cl _3$
  • B
    $\left[ Cr \left( NH _3\right)_4 Cl _2\right] Cl$
  • $\left[ Cr \left( NH _3\right)_5 Cl \right] Cl _2$
  • D
    $\left[ Cr \left( NH _3\right)_3 Cl _3\right]$

Answer

Correct option: C.
$\left[ Cr \left( NH _3\right)_5 Cl \right] Cl _2$
(C)
$\begin{array}{c}\text { Given : } \Delta T _{ f }=0.558^{\circ} C \\ k _{ f }=1.86 \frac{K \times kg }{ mol } \\ 0.1 m \text { aq. sol. } \\ \Rightarrow \Delta T _{ f }= i \times k _{ f } \times m \\ \Rightarrow 0.558= i \times 1.86 \times 0.1 \\ \Rightarrow i =3\end{array}$

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