A certain mass of a substance when dissolved in `100 g C_(6)H_(6)` lowers the freezing point by `1.28^(circ)C`. The same mass of solute dissolved in `100 g` of water lowers of the freezing point by `1.40^(circ)C`. If the substance has normal molecular weight in benzene and is completely dissocited in water, into how many ions does it dissocite in water ? `K_(f)` for `H_(2)O` and `C _(6)H_(6)` are `1.86` and `5.12 K mol^(-1) kg` respectively.

Answer

3 ions

Explanation

Solute-B, solvent-A
In benzene substance does not dissociate
`Delta"T"_("f")=("K"_("f")xx"W"_("B")xx1000)/("m"_("B")xx"W"_("B"))`
`1.28=(5.12xx"W"_("B")xx1000)/("m"_("B")xx100)`………….(i)
In water substance dissociation completely (say into'n' ions)
`therefore"i=n"`
`Delta"T"_("f")^(1)="i"xx"K"_("f")^(1)("K"_("f")^(1)xx"W"_("B")^(0)xx1000)/("m"_("B")^(0)xx"W"_("A")^(0))`
`1.40="n"xx(1.86xx"W"_("B")xx1000)/("m"_("B")xx100)`..........(ii)
On dividing equation (ii) by equation (i)
`1.40/1.28=(1.86"n")/(5.12)`
`"n"=(1.40xx5.12)/(1.28xx1.86)`
`"n"=3`
`therefore"substance dissociates into three ions"`