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(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273.16 \ K - 272.814 \ K = 0.346 \ K$. The formula for depression i

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$50$

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(D) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 273.16 \ K - 272.814 \ K = 0.346 \ K$. The formula for depression in freezing point is $\Delta T_f = K_f 	imes m$,where $m$ is the molality. Molality $m = \frac{w_A 	imes 1000}{M_A 	imes w_{solvent}}$,where $w_A = 0.2 \ g$ and $w_{solvent} = 21.5 \ g$. Substituting the values: $0.346 = 1.86 	imes \frac{0.2 	imes 1000}{M_A 	imes 21.5}$. $M_A = \frac{1.86 	imes 0.2 	imes 1000}{0.346 	imes 21.5} = \frac{372}{7.439} \approx 50 \ g \ mol^{-1}$. Thus,the molar mass of solute '$A$' is $50 \ g \ mol^{-1}$.

An aqueous solution containing $0.2 \ g$ of a non-volatile solute '$A$' in $21.5 \ g$ of water freezes at $272.814 \ K$. If the freezing point of water is $273.16 \ K$,the molar mass (in $g \ mol^{-1}$) of solute '$A$' is $[K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1}]$

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