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A `5%` solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The freezing point of pure water is 273.15 K.
A. `271 K`
B. `273.15K`
C. `269.07K`
D. `277.23K`

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Correct Answer - C
`DeltaT_(f)=(1000xxK_(f)xxW_(2))/(W_(1)xxM_(2))`
`DeltaT_(f)` (cane sugar) `= 273.15-271=2.15K`
`therefore2.15=(1000xxK_(f)xx5)/(95xx342)`……(i)
`DeltaT_(f)("glucose")=(1000xxK_(f)xx5)/(95xx180)`…..(ii) Dividing (ii) by (i)
`(DeltaT_(f)("glucose"))/(2.15)=(342)/(180)or DeltaT_(f)("glucose")=4.08^(@)`
`therefore` freezing point glucose solution
`=273.15-4.08=269.07K`

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