Question

A sample of industrial waste water is found to contain 8.2% $N{a}_{3}P{O}_4$ and 12% $MgS{O}_4$ by weight in solution. If % ionisation of $N{a}_{3}P{O}_4$ and $MgS{O}_4$ are 50% and 60% respectively, then its normal boiling point (in °C ) is (2 decimal places).
[use $K_b\left(H_{2}O\right)$ = 0.50 $Kkgmo{1}^{-}1$]:

Answer 1

$\mathrm{For}{\mathrm{Na}}_3{\mathrm{PO}}_{4,}i=1+3\alpha =1+3\times 0.5=2.5\mathrm{For}{\mathrm{MgSO}}_4,i=1+\alpha =1+0.6=1.6$
$\mathrm{Let},100g\mathrm{of}\mathrm{water}\mathrm{contain}8.2g{\mathrm{Na}}_3{\mathrm{PO}}_4\mathrm{and}12g{\mathrm{MgSO}}_4\Delta T_b=K_b\mathrm{mi}=K_b[\frac{\mathrm{effective}\mathrm{no}.\mathrm{of}\mathrm{moles}\mathrm{of}({\mathrm{Na}}_3{\mathrm{PO}}_4+{\mathrm{MgSO}}_{4)}}{\mathrm{wt}.\mathrm{of}\mathrm{solvent}\left(\mathrm{in}\mathrm{gm}\right)}\times 1000]\Delta T_b=0.50[\frac{\frac{8.2}{164}\times 2.5+\frac{12}{120}\times 1.6}{79.8}\times 1000]=1.785$
$T_b=100+1.785=101.785°C$