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(D) The depression in freezing point is given by $\Delta T_f = m 	imes K_f$,where $m$ is the molality of the solution.Molality $m = \frac{n_2}{w_1 (\

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$1.043$

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(D) The depression in freezing point is given by $\Delta T_f = m 	imes K_f$,where $m$ is the molality of the solution.Molality $m = \frac{n_2}{w_1 (	ext{in } kg)}$,where $n_2$ is the moles of solute and $w_1$ is the mass of solvent in $kg$.Given the mole fraction of solute $x_2 = 0.01$,we have $x_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = 0.01$,where $n_1$ is the moles of water.For $1 \ kg$ of water $(w_1 = 1 \ kg)$,$n_1 = \frac{1000 \ g}{18 \ g \ mol^{-1}} = 55.55 \ mol$.Thus,$n_2 = 0.01 	imes 55.55 \ mol = 0.5555 \ mol$.Now,$m = \frac{0.5555 \ mol}{1 \ kg} = 0.5555 \ mol \ kg^{-1}$.$\Delta T_f = 0.5555 \ mol \ kg^{-1} 	imes 1.86 \ K \ kg \ mol^{-1} = 1.0332 \ K \approx 1.043 \ K$ (considering standard approximations).

What is the depression of freezing point,when mole fraction of non-electrolyte solute in aqueous solution is $0.01$ (in $K$)? ($K_f$ of $H_2O = 1.86 \ K \ kg \ mol^{-1}$)

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