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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 mass of solute

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 mass of solute is $5 \ g$ and the mass of solvent is $95 \ g$.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$.Since $w_2$ and $w_1$ are the same for both cane sugar $(M_2 = 342 \ g/mol)$ and glucose $(M_2 = 180 \ g/mol)$,$\Delta T_f$ is inversely proportional to the molar mass $M_2$.$\Delta T_{f, 	ext{sugar}} = 273.15 - 271 = 2.15 \ K$.$\frac{\Delta T_{f, 	ext{glucose}}}{\Delta T_{f, 	ext{sugar}}} = \frac{M_{	ext{sugar}}}{M_{	ext{glucose}}} = \frac{342}{180} = 1.9$.$\Delta T_{f, 	ext{glucose}} = 1.9 	imes 2.15 = 4.085 \ K$.Freezing point of glucose solution $= 273.15 - 4.085 = 269.065 \ K \approx 269.07 \ K$.

If the freezing point of a $5\%$ (by mass) aqueous solution of cane sugar is $271 \ K$ and the freezing point of pure water is $273.15 \ K$,then the freezing point of a $5\%$ (by mass) aqueous solution of glucose will be .......... $K$.

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