1.5 g of `Ba(NO_(3))_(2)` dissolved in 100g of water shows a depression in freezing point equal to `0.28^(@)C` . What is the percentage dissociation of the salt ? (`k_(f)` for water ` = 1.86 K//m` and molar mass of `Ba(NO_(3))_(2) = 261`).

Answer

`81%`

Explanation

`M_(B) ` (observed) `= (1.86 xx 1000xx 1.5)/(0.280 xx 100)= 99.64`
` i=(261)/(99.64) = 2.62`
`Ba(NO_(3))_(2) hArr Ba^(2+) + 2NO_(3)^(-)`
`{:(,"Initial",1,0,0),(,"After dissociaiton ",1-alpha,alpha,2alpha):}`
` i=(1-alpha + alpha+2alpha)/(1) = 2.62`
`alpha = 81%`