The freezing point of a 5% (by mass) solution | vedclass

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(C) For cane sugar solution: $(\Delta T_f)_{cane} = K_f 	imes m = K_f 	imes \frac{5 	imes 1000}{342 	imes 100} = 273.15 - 271 = 2.15 \, K$ For glu

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$269.07$

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(C) For cane sugar solution: $(\Delta T_f)_{cane} = K_f 	imes m = K_f 	imes \frac{5 	imes 1000}{342 	imes 100} = 273.15 - 271 = 2.15 \, K$ For glucose solution: $(\Delta T_f)_{glucose} = K_f 	imes m = K_f 	imes \frac{5 	imes 1000}{180 	imes 100}$ Dividing the two equations: $\frac{(\Delta T_f)_{glucose}}{2.15} = \frac{342}{180}$ $(\Delta T_f)_{glucose} = 2.15 	imes \frac{342}{180} = 4.085 \, K$ Freezing point of glucose solution = $273.15 - 4.085 = 269.065 \, K \approx 269.07 \, K$.

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

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