2 g of a non-electrolyte solute (molar mass | vedclass

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(B) 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

$0.3$

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(B) 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: $w_1 = 2 \ g$,$m_1 = 500 \ g \ mol^{-1}$,$w_2 = 57.3 \ g$,$K_f = 4.3 \ K \ kg \ mol^{-1}$. Substituting the values: $\Delta T_f = 4.3 	imes \frac{2 	imes 1000}{500 	imes 57.3}$. $\Delta T_f = 4.3 	imes \frac{2000}{28650} = 4.3 	imes 0.0698 \approx 0.3 \ K$.

$2 \ g$ of a non-electrolyte solute (molar mass is $500 \ g \ mol^{-1}$) was dissolved in $57.3 \ g$ of xylene. If the freezing point depression constant $K_f$ of xylene is $4.3 \ K \ kg \ mol^{-1}$,then the depression in freezing point of xylene is.......... (in $K$)

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