Question

An 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 is:
$\left[K_b\right(H_{2}O)=0.50Kkgmo{l}^{-1}$]

• A. ${102.3}^{\circ }C$
• B. ${103.35}^{\circ }C$
• C. ${101.785}^{\circ }C$
• D. None of these

Answer 1

The correct option is C ${101.785}^{\circ }C$

For $N{a}_{3}P{O}_4$,
$i=1+3\alpha =1+3\times \frac{50}{100}=2.5$

For $MgS{O}_4$,
$i=1+\alpha =1+\frac{60}{100}=1.6$

$100g\text{solution contains}8.2gN{a}_{3}P{O}_{4}and12gMgS{O}_4.$

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b[\frac{\text{Effective no. of moles of}(N{a}_{3}P{O}_4+MgS{O}_4)}{\text{Weight of solvent (in g)}}\times 1000]$

$\Delta T_b=0.50\left[\frac{\frac{8.2}{164}\times 2.5+\frac{12}{120}\times 1.6}{79.8}\right]\times 1000={1.785}^{\circ }C$

$T_b=100+1.785={101.785}^{\circ }C$