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 $\%$ ionization of $N{a}_{3}P{O}_4$ and $MgS{O}_4$ are $50$ and $60$ respectively then it’s normal boiling point (in ${}^{o}C$) is: $\left[k_b\right(H_{2}O)=0.50{\text{K kg mol}}^{-1}$]:

Answer 1

$$\begin{array}{ccc}N{a}_{3}P{O}_4& \to 3N{a}^{+}& +P{O}_4^{3-}\\ 1-\alpha & 3\alpha & \alpha \end{array}$$
$i=1+3\alpha$
$i=1+3\times \frac{50}{100}=2.5$
$$\begin{array}{ccc}MgS{O}_4& \to M{g}^{2+}& +S{O}_4^{2-}\\ 1-\alpha & \alpha & \alpha \end{array}$$
$i=1+\alpha$
$i=1+\frac{60}{100}=1.6$
According to given data $100\text{g}$ solution contain $8.2\text{g}N{a}_{3}P{O}_4$ and $12\text{g}MgS{O}_4$
Mass of solvent $=100–(8.2+12)=79.8\text{g}$
(Molar mass of $N{a}_{3}P{O}_4=164\text{g/mol}$)
(Molar mass of $MgS{O}_4=120\text{g/mol}$)
$\Delta T_b=i\times k_b\times m$
$=\frac{\text{Effective number of moles}(N{a}_{3}P{O}_4+MgS{O}_4)}{\text{Weight of solvent in kg}}\times 1000$
$=0.50\left[\frac{\frac{8.2}{164}\times 2.5+\frac{12}{120}\times 1.6}{79.8}\right]\times 1000$
$=0.50\left[\frac{0.125+0.16}{79.8}\right]\times 1000$
$\Delta T_b={1.785}^{o}C$
$\Delta T_b=\text{Boiling point of solution}\left(T_b\right)-\text{Boiling point of solvent}\left(T_b^o\right)$
$T_b=100+1.785$
$T_b={101.78}^{o}C$