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

`DeltaT=(1000xxK_(f)xxw_(2))/(w_(1)xxM_(2))`
As substance has normal molecular mass in benzene, when dissolved in benzene, we have
`1.28=(1000xx5.12xxw_(2))/(100xxM_("normal"))" …(i)"`
As substance dissociates in water, observed molecular mass can be calculated as
`1.40=(1000xx1.86xxw_(2))/(100xxM_("observed"))" ...(ii)"`
Dividing eqn. (ii) by eqn. (i)
`(M_("normal"))/(M_("observed"))=(1.40)/(1.28)xx(5.12)/(1.86)=3.01" "therefore" "i=3.01~=3.0`
As the substance (solute) is `100%` ionized, `alpha=1.`
Suppose the solute is `A_(x)B_(y)`
`{:(,A_(x)B_(y),hArr,xA^(+),+,yB^(-),),("Initial moles ",1,,0,,0,),("After disso.",1-alpha,,xalpha,,yalpha","," Total "=1-alpha+xalpha+yalpha):}`
`therefore" "i=1-alpha+x alpha+yalpha`
`therefore" "3=1-1+x+y" " (because i=3, alpha=1)`
`"or "x+y=3`
`therefore" No. of ions produced on dissociation = 3"`