The freezing point of 0.1 M solution of glucose is `-1.86^(@)C`. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be

Options

A. `-7.44^(@)C`
B. `-5.58^(@)C`
C. `-3.72^(@)C`
D. `-2.79^(@)C`

Answer

C

Explanation

`K_(1)=DeltaT_(f)//M=1.86//0.1=18.6`
`0.1xxV+0.3V=M_(3)xx2VorM_(3)=0.2,`
`DeltaT_(f)=18.6xx0.2=3.72^(@)C`
`=-3.72^(@)C`
`DeltaT_(f)=T_(f)^(@)-T_(f)=0^(@)-T_(f)`
`T_(f)=0-3.72=-3.72^(@)C`