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))`

Options

A. (a)0.32
B. (b)3.2
C. (c )0.032
D. (d)0.48

Answer

B

Explanation

Mole fraction of `I_(2)=X_(Az)`
`X_(A)=n_(1)/(n_(1)+n_(2)) X_(B)=n_(2)/(n_(1)+n_(2))`
`X_(A)+X_(B)=1 " " X_(A)=0.2`
`X_(B)=0.8" " X_(A)/X_(B)=n_(1)/n_(2)`
`n_(1):n_(2)=1:4`
`(("Mass of solvent",,,,),( :'n_(2)=w/M,,,,),(w=n_(2)xxM=4xx78(g),,,,):}`
`"Molality"= n_(1)/("mass of solvent" ("in kg"))` `=1/(4xx78)xx1000=3.21 m`