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(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$ (since $NaCl ight

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$1.62$

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(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$ (since $NaCl ightarrow Na^+ + Cl^-$). Given: $\Delta T_f = 6 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and mass of solvent $W = 1 \ kg$. Substituting the values: $6 = 2 	imes 1.86 	imes m$. $m = \frac{6}{2 	imes 1.86} = \frac{6}{3.72} \approx 1.6129 \ mol \ kg^{-1}$. Since the mass of solvent is $1 \ kg$,the amount of $NaCl$ is approximately $1.61 \ mol$ (rounded to $1.62 \ mol$ based on options).

The depression in freezing point of a $NaCl$ solution is $6 \ K$. If $K_f = 1.86 \ K \ kg \ mol^{-1}$ for water,then the amount of $NaCl$ dissolved in $1 \ kg$ of water is ............. $mol$?

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