A solution containing 3.100 g of `BaCl_(2)` in 250 g of water boils at `100.083^(@)C` .Calculate the Van't Hoff factor and molality of `BaCl_(2)` in this solution . (`k_(b)` for water `= 0.52 Km^(-1)` , molar mass of `BaCl_(2) = 208.3 g mol^(-1)`)

Explanation

Molality of the solution ,
`m= (w_(B) xx 1000)/(M_(B) xx w_(A))`
`w_(B) = 3.100 g, w_(A) = 250g , M_(B) = 208.3`
`m = (3.100 xx 1000)/(208.3 xx 250) = 0.05952`
Now , let us calculate normal elevation in boiling point ,
`Delta T_(b) = k_(b) xx m`
` = 0.05952 xx 0.52 = 0.03095`
Observed elevation in boiling point,
`Delta T_(b) = 100.083 - 100 = 0.083^(@)C`
` i=("Observed" Delta T_(b))/("Normal" Delta T_(b)) = (0.083)/(0.03095) = 2.68`