1.

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.

Answer»

271 K
273.15
269.07 K
277.23 K.

Solution :For both the solution `W_(B)andW_(A)` are the same.
`(DeltaT_(f(g)))/(DeltaT_(f(s)))=(M_(("suorse")))/(M_(("GLUCOSE")))`
`DeltaT_(f(g))=((342g MOL^(-1)))/((182g mol^(-1)))xx215K=4.04 K`
`"f.p. of glucose solution"=(273.15-4.-4)`
=269.11 K.


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