An aqueous solution containing 12.48 g of barium chloride in 1.0 kg of water boils at 373.08321K. Calculate the d.egree of dissociation of barium I chloride. (Given `K_(b)`, of `H_(2)O = 0.52 K m^(-1)`, molar mass of Ba `CI_(2) = 208.34 g mol^(-1)`]

Explanation

`W_(R) = 12.48 ` g of `BaCI_(2)`
`W_(A) = 1 `kg
Boiling point of solution = 373.0832 K
Degree of dissociation of `BaCI_(2)`= ?
`K_(b)=0.52 km^(-1)`
`M_(B)= 208.34` g/mol .
Boiling point of water =373 K
`T_(b) = iK_(b)m`
`0.0832 = ixx0.52xx(12.48)/(208.34) xx1 kg `
`i= (208.34xx0.0832)/(0.52xx12.48)`
`=(17.33)/(6.48)`
i = 2.67
`BaCI_(2)= Ba^(2+)+ 2CI^(-) `
1 mol `" "` 0 `" "` 0
`1-alpha " " alpha " " 2 alpha`
Total 1 + 2 `alpha`
`i =1 +2 alpha ` or `alpha = (i-1)/(2) = (2.67-1)/(2)`
`= (1.67)/(2) = 0.835`