A 0.01 M solution of glucose in water freezes at `-0.0186^(@)C` A 0.01 M solution of `KNO_(3)` in water will freeze at

Options

A. `-0.0093^(@)C`
B. `-0.0372^(@)C`
C. `-0.0186^(@)C`
D. `-0.093^(@)C`

Answer

B

Explanation

`KNO_(3)` ionises as,
`KNO_(3) hArr K^(+) + NO_(3)^(-)`
`therefore` Concentration in solution `=2xx0.01=0.02`
`DeltaT_(f)=T_(f)-T^(@)=K_(f)xxm` for glucose
`therefore k_(f)(DeltaTf)/(m)=(0.018)/(0.01)=1.86^(@) m^(-1)`
`DeltaT_(f)` for `KNO_(3)=1.86xx0.02=0.0372^(@)C`
`T_(f)=T^(@)-T_(f)=0.0000-0.0372=-0.0372^(@)C`