The freezing point of a 5% (by mass) aqueous | 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 $5\%$ (by mass) solutions,the mass of solut

Vedclass · Accepted Answer

$269.07$

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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 $5\%$ (by mass) solutions,the mass of solute is $5 \, g$ and the mass of solvent is $95 \, g$.$\Delta T_f = K_f 	imes \frac{w_2 	imes 1000}{M_2 	imes w_1}$.Since $w_2$,$w_1$,and $K_f$ are the same for both,$\Delta T_f \propto \frac{1}{M_2}$.$\Delta T_f(	ext{sucrose}) = 273.15 - 271 = 2.15 \, K$.$\frac{\Delta T_f(	ext{glucose})}{\Delta T_f(	ext{sucrose})} = \frac{M(	ext{sucrose})}{M(	ext{glucose})} = \frac{342}{180} = 1.9$.$\Delta T_f(	ext{glucose}) = 2.15 	imes 1.9 = 4.085 \, K$.Freezing point of glucose solution $= 273.15 - 4.085 = 269.065 \, K \approx 269.07 \, K$.

The freezing point of a $5\%$ (by mass) aqueous solution of sucrose is $271 \, K$ and the freezing point of pure water is $273.15 \, K$. The freezing point of a $5\%$ (by mass) aqueous solution of glucose is .......... $K$.

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