For Na3PO4, i=1+3α =1+3×0.5=2.5For MgSO4, i=1+α =1+0.6=1.6
Let, 100g of water contain 8.2g Na3PO4 and 12g MgSO4ΔTb= Kbmi =Kb[effective no. of moles of (Na3PO4+MgSO4)wt. of solvent (in gm)×1000] ΔTb=0.50[8.2164×2.5+12120×1.679.8×1000]=1.785
Tb=100+1.785=101.785°C