1.

A solutioncontaining 3.100 g of BaCl_(2) in 250 gof waterboilsat 100.083^(@)C .Calculate the Van't Hoff factorand molality of BaCl_(2) in thissolution . (k_(b) for water = 0.52 Km^(-1) , molar mass of BaCl_(2) = 208.3 g mol^(-1))

Answer»

Solution :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 elevationin 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`


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