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(D) The depression in freezing point is given by the formula: $\Delta T_f = i \cdot K_f \cdot m$Here,$\Delta T_f = 0 - (-0.00732) = 0.00732 \ ^oC$,$K_

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$2$

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(D) The depression in freezing point is given by the formula: $\Delta T_f = i \cdot K_f \cdot m$Here,$\Delta T_f = 0 - (-0.00732) = 0.00732 \ ^oC$,$K_f = 1.86 \ ^oC/m$,and $m = 0.0020 \ m$.Calculating the van't Hoff factor $(i)$:$i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.00732}{1.86 	imes 0.0020} = \frac{0.00732}{0.00372} \approx 1.968 \approx 2$.The ionic compound $[Co(NH_3)_5(NO_2)]Cl$ dissociates in water as: $[Co(NH_3)_5(NO_2)]Cl ightarrow [Co(NH_3)_5(NO_2)]^+ + Cl^-$.Thus,$1 \ mol$ of the compound produces $2 \ mol$ of ions.

$A$ $0.0020 \ m$ aqueous solution of an ionic compound $[Co(NH_3)_5(NO_2)]Cl$ freezes at $-0.00732 \ ^oC.$ Number of moles of ions which $1 \ mol$ of ionic compound produces on being dissolved in water will be
$(K_f = 1.86 \ ^oC/m)$

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$A$ $0.0020 \ m$ aqueous solution of an ionic compound $[Co(NH_3)_5(NO_2)]Cl$ freezes at $-0.00732 \ ^oC.$ Number of moles of ions which $1 \ mol$ of ionic compound produces on being dissolved in water will be $(K_f = 1.86 \ ^oC/m)$