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Mole fraction of I(2) in C(6) H(6) is 0....

Mole fraction of `I_(2)` in `C_(6) H_(6)` is 0.2. Calculate molality of `I_(2)` in `C_(6) H_(6)`. `(Mw of C_(6) H_(6) = 78 g mol^(-1))`

A

`3.2`

B

`6.40`

C

`1.6`

D

`2.30`

Text Solution

Verified by Experts

The correct Answer is:
A
`(n)/(n+N) = 0.2 :. (N)/(n+N) = 0.8`
or `(n)/(N) = (1)/(4)`
or `(n_(I_(2)))/(w_(C_(6)H_(6))) xx M_(C_(6)H_(6)) = (1)/(4)`
`:.` Molality `=(n_(I_(2)))/(w_(C_(6)H_(6))) xx 1000`
`= (1)/(4) xx (1000)/(M_(C_(6)H_(6)))`
or Molality `= (1)/(4) xx (1000)/(78)`
`= 3.2`
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