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% so

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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% so

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.

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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.

8 views Answered 2023-02-05 15:29