1.00 g of a non-electrolyte solute (molar ma | vedclass

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(A) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{W_1 	imes

Vedclass · Accepted Answer

$0.4$

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(A) The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{W_1 	imes 1000}{M_1 	imes W_2}$,where $W_1$ is the mass of solute,$M_1$ is the molar mass of solute,and $W_2$ is the mass of solvent in grams.Given: $W_1 = 1.00 \ g$,$M_1 = 250 \ g \ mol^{-1}$,$W_2 = 51.2 \ g$,and $K_f = 5.12 \ K \ kg \ mol^{-1}$.Substituting the values: $\Delta T_f = 5.12 	imes \frac{1.00 	imes 1000}{250 	imes 51.2}$.$\Delta T_f = 5.12 	imes \frac{1000}{12800} = 5.12 	imes 0.078125 = 0.4 \ K$.

$1.00 \ g$ of a non-electrolyte solute (molar mass $250 \ g \ mol^{-1}$) is dissolved in $51.2 \ g$ of benzene. If the freezing point depression constant of benzene is $5.12 \ K \ kg \ mol^{-1}$,then the freezing point of benzene decreases by ...... $K$.

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