Question

KBr is `180%` dissociated in aqueous solution of `0.5m` concentration. (Given,`K_(f)"for water "=1.86Kkg" mol"^(-1)` ). The solution freezes at
A. `271.326K`
B. 272 K
C. `270.5K`
D. `268.5K`

Answer 1

Correct Answer - A
`DeltaT_(f)=ixxK_(f)xx "molality"`
Here,degree of dissociation `=80%`
`becausei=(1-alpha+nalpha)/(1)`
`KBrhArrK^(+)+Br^(-)`
`i=(1-0.8+2xx0.8)/(1)=1.8`
`:.DeltaT_(f)=1.8xx1.86xx0.5=1.674K`
`DeltaT_(f)^(@)-T_(f)rArr1.678=273-T_(f)`
`T_(f)=271.326K`