A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound…

A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_{3}ight)_{5}\left(\mathrm{NO}_{2}ight) \mathrm{Cl}$ 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 $\left(\mathrm{K}_{\mathrm{f}}=-1.86{ }^{\circ} \mathrm{C} / \mathrm{m}ight)$
  1. 2
  2. 1
  3. 3
  4. 4

Solution

$\Delta \mathrm{T}_{\mathrm{f}}=0-\left(-0.00732^{\circ}ight)=0.00732$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \mathrm{m}$
$\mathrm{i}=\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}} \times \mathrm{m}}=\frac{0.00732}{1.86 \times 0.002}=2$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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