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.

Options

A. 271 K
B. 273.15 K
C. 269.07 K
D. 277.23 K

Answer

c

Explanation

`K_(f)` for water `=(Delta T_(f)xxWxxm)/(1000xxw)`
( where W= wof water w= wt of can sugar, m= molecular wt. of cane sugar)
`=(2.15xx100xx342)/(1000xx5)=14.7`
Now, for 5% glucose,
`DeltaT_(f) =(K_(f)xx1000xxw)/( Wxxm')` ( where wt. of glucose, m= molecular, wt. of glucose)
`=(14.7xx1000xx5)/(100xx180)=4.08`
`:.` Freezing point of glucose solution
`=213.15-4.08=269.07 K`