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(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{

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$1.5$

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(B) The formula for the depression in freezing point is $\Delta T_{f} = K_{f} 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$,where $W_2$ is the mass of the solute,$M_2$ is the molar mass of the solute,and $W_1$ is the mass of the solvent in grams.Substituting the given values: $W_2 = 3.2 \ g$,$M_2 = 128 \ g \ mol^{-1}$,$W_1 = 80 \ g$,and $K_{f} = 4.8 \ K \ kg \ mol^{-1}$.$\Delta T_{f} = \frac{4.8 \ K \ kg \ mol^{-1} 	imes 3.2 \ g 	imes 1000 \ g \ kg^{-1}}{128 \ g \ mol^{-1} 	imes 80 \ g}$.$\Delta T_{f} = \frac{15360}{10240} \ K = 1.5 \ K$.

Find the depression in freezing point of a solution when $3.2 \ g$ of a non-volatile solute with a molar mass of $128 \ g \ mol^{-1}$ is dissolved in $80 \ g$ of solvent,given that the cryoscopic constant of the solvent is $4.8 \ K \ kg \ mol^{-1}$. (in $K$)

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