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A 5% solution (by mass) of cane sugar in...

A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.

A

271 K

B

273.15 K

C

269.07 K

D

277.23 K

Text Solution

Verified by Experts

The correct Answer is:
c
`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`
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