Exam

NEET

Year

2009

Subject

Chemistry

Chapter

Solutions

NEET 2009 · Chemistry · Solutions

A 0.0020 ~m aqueous solution of an ionic compound Co(NH_3)_5(NO_2)...

A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which $1 \mathrm{~mol}$ of ionic compound produces on being dissolved in water will be $\left(\mathrm{k}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$

Step-by-step solution

Given, $$\begin{aligned}\text{ molality, } \mathrm{m}& =0.0020 \mathrm{ m}\\ \Delta \mathrm{T}_{\mathrm{f}}& =0^{{}^{\circ}} \mathrm{C}-0.00732^{{}^{\circ}} \mathrm{C}=0.00732^{{}^{\circ}} \mathrm{C}\\ \mathrm{k}_{\mathrm{f}}& =1.86^{{}^{\circ}} \mathrm{C}/ \mathrm{m}\\ \Delta \mathrm{T}_{\mathrm{f}}& =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m}\\ \mathrm{i}& =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}}\\ & =\frac{0.00732}{1.86 \times 0.0020}\\ & =1.96 \approx2 \end{aligned}$$Since,$the compound is ionic, so number of moles produced is equal to vant' Hoff factor, i$. Hence, 2 moles of ions are produced.
$