1.

A solution containing 3.1 g of BaCI_(2)in this solution. (K_(b) for water= 0.52 K m^(-1). Molar mass BaCI_(2)=208.3 mol^(-1)).

Answer»


Solution :`"Molality of solution (m)"=W_(B)/(M_(B)xxW_(A))`
`W_(B)=3.1g, W_(A)=250 g =0.25 kg, M_(B)=208.3 " g MOL"^(-1)`
`m=((3.1g))/((208.3" g mol"^(-1))xx(0.25 kg))=0.05952" kgmol"^(-1)=0.05952 m.`
`"For the solution", Delata T_(b)=iK_(b)m`
`or"" i=(DeltaT_(b))/(K_(b)xxm),DeltaT_(b)=100.083^(@)C-100^(@)C=0.083 K`
`K_(b)=0.52Km^(-1),m=0.05952 m`
`THEREFORE""i=((0.083 K))/((0.52"Km"^(-1))xx(0.05952m))=2.68`
`"RATIO of i and m (i/m)"=(2.68)/(0.05952)=45:1`


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