$\mathrm{CrCl_3\cdot xNH_3}$ 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 $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$)

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\Delta T_f=iK_fm\Rightarrow i=\frac{0.558}{1.86	imes0.1}=3$. Three particles per formula unit means one complex ion and two free chloride ions: $[\mathrm{Cr(NH_3)_5Cl}]\mathrm{Cl_2}$ (coordination number 6). Answer (1).

Q. $\mathrm{CrCl_3\cdot xNH_3}$ 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 $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$)

Explanation

Reviewed by Dr. Abubacker

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