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.

Explanation

`Delta T_(f) ` is given as :
`Delta T_(f) = (k_(f) xx 1000 xx w_(B))/( M_(B) xx 100)`
Since the given solute behave as normal in `C_(6)H_(6) ` and dissociated in water .
` i=("M(normal)")/("M(Observed)")= ("M(in benzene)")/("M(in water)")`
In `C_(6)H_(6) " " 1.06 = (5.12 xx 1000 xx w_(B))/(M("in benzene") xx 100)" "........(i)`
In water `" " 1.15 = (1.86 xx 1000 xx w_(B))/("M(in water )" xx 100) " " .......(ii)`
Dividing equation (ii) by eq.(i)
`("M(in benzene)")/("(M(in water)") = (1.15)/(1.06) xx (5.12)/(1.86) = 2.99`
or `" " i=2.99 ~~3`
Sine the solute is completely dissociated, `alpha = 1` , the solute is ionised into three ions in water. If solute is `A_(x)B_(y)` .
`A_(x) B_(y) = xA^(+) + yB^(-)`
` x + y = 3` .