1 . 5 of Ba `(NO_(3))_(2)` dissolved in 100 g 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

`W_(B) = 1 . 50 g, M_(B) = 261 g mol^(-1)`
`W_(A) = 100 g = 0 . 1 kg`
`K_(f) = 1 . 86 Km^(-1)`
`= 1 . 86 K kg mol^(-1)`
`Delta T_(f) = 0 . 28 ^(@) C = 0 . 28 K, i = ?`
`Delta T_(f) = i xx K_(f) xx (W_(B))/( M_(B)) xx (1)/( W_(A) (kg))`
`i = (Delta T_(f) xx M_(B) xx W_(A) (kg))/( K_(f) xx W_(B))`
`= ((0 . 28 K) xx (261 g mol^(-1)) xx ( 0 . 1 kg))/( ( 1 . 86 K kg mol^(-1)) xx 1 . 50 g)`
` = (18 xx 261)/( 186 xx 15) = 2 . 62 `
`Ba (NO_(3))_(2) (aq) hArr Ba^(2+) (aq) + 2 NO_(3)^(-) (aq)`
`{:("Initial Cone 1 ___ ___ "),("At . eqn . " 1 - alpha " "alpha " "2 alpha ):}`
`i = (1 - alpha + alpha + 2 alpha)/( 1) = 1 + 2 alpha`
`:. 1 + 2 alpha = 2 . 62`
`2 alpha = 2 . 62 - 1 = 1 . 62`
`alpha = (1 . 62)/( 2) = 0 . 81`
Percentage dissociation ` = 100 xx alpha `
`= 100 xx 0 . 81`
`= 81% `