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

A certain mass of a substance when dissolved in 100 g C_(6)H_(6) lower

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