A 0.0020 m0.0020 \mathrm{~m}0.0020 m aqueous solution of an ionic compound Co(NH3)5(NO2)Cl\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}Co(NH3)5(NO2)Cl freezes at −0.00732∘C-0.00732^{\circ} \mathrm{C}−0.00732∘C. Number of moles of ions which 1 mol1 \mathrm{~mol}1 mol of ionic compound produces on being dissolved in water will be (kf=−1.86∘C/m)\left(\mathrm{k}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C} / \mathrm{m}\right)(kf=−1.86∘C/m)
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Given, molality, m=0.0020 mΔTf=0∘C−0.00732∘C=0.00732∘Ckf=1.86∘C/mΔTf=i⋅kf×mi=ΔTfkf×m=0.007321.86×0.0020=1.96≈2\begin{aligned} \text { molality, } \mathrm{m} & =0.0020 \mathrm{~m} \\ \Delta \mathrm{T}_{\mathrm{f}} & =0^{\circ} \mathrm{C}-0.00732^{\circ} \mathrm{C}=0.00732^{\circ} \mathrm{C} \\ \mathrm{k}_{\mathrm{f}} & =1.86^{\circ} \mathrm{C} / \mathrm{m} \\ \Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\ \mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}} \\ & =\frac{0.00732}{1.86 \times 0.0020} \\ & =1.96 \approx 2 \end{aligned} molality, mΔTfkfΔTfi=0.0020 m=0∘C−0.00732∘C=0.00732∘C=1.86∘C/m=i⋅kf×m=kf×mΔTf=1.86×0.00200.00732=1.96≈2 Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, iii. Hence, 2 moles of ions are produced.
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