Question

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

Answer 1

Correct Answer - c
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.