A 5% solution (by mass) of cane sugar in wate | vedclass

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(C) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality.For a $5\%$ solution (by mass),the molality $m

Vedclass · Accepted Answer

$269.07$

Vedclass · Answer

(C) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality.For a $5\%$ solution (by mass),the molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$.Since the percentage by mass is the same $(5\%)$,the molality is inversely proportional to the molar mass $(M_2)$ of the solute.$\Delta T_f \propto \frac{1}{M_2}$.For cane sugar $(M_1 = 342 \ g/mol)$: $\Delta T_{f1} = 273.15 - 271 = 2.15 \ K$.For glucose $(M_2 = 180 \ g/mol)$: $\Delta T_{f2} = \Delta T_{f1} 	imes \frac{M_1}{M_2}$.$\Delta T_{f2} = 2.15 	imes \frac{342}{180} = 2.15 	imes 1.9 = 4.085 \ K$.The freezing point of the glucose solution $= 273.15 - 4.085 = 269.065 \ K \approx 269.07 \ K$.

$A$ $5\%$ solution (by mass) of cane sugar in water has a freezing point of $271 \ K$ and the freezing point of pure water is $273.15 \ K$. The freezing point of a $5\%$ solution (by mass) of glucose in water is ............. $K$.

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