Solutions and Colligative Properties 2 Question 27
27. The vapour pressure of pure benzene is $639.70 \mathrm{~mm}$ of $\mathrm{Hg}$ and the vapour pressure of solution of a solute in benzene at the same temperature is $631.9 \mathrm{~mm}$ of $\mathrm{Hg}$. Calculate the molality of the solution.
(1981, 3M)
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Solution:
- From the graph we can note
$\Delta T_{b}$ for solution $X$ i.e.,
$$ \Delta T_{b(X)}=362-360=2 $$
Likewise, $\Delta T_{b}$ for solution $Y$ i.e., $\Delta T_{b(Y)}=368-367=1$
Now by using the formula
$$ \Delta T_{b}=i \times \text { molality of solution } \times K_{b} $$
For solution $X$
$$ 2=i \times m_{\mathrm{NaCl}} \times K_{b(X)} $$
Similarly for solution $y$
$$ 1=i \times m_{\mathrm{NaCl}} \times K_{b(Y)} $$
from Eq. (i) and (ii) above
$$ \frac{K_{b(X)}}{K_{b(Y)}}=\frac{2}{1} \text { or } 2 \quad \text { or } \quad K_{b(X)}=2 K_{b(Y)} $$
For solute $S$
$$ \begin{array}{lcc} & 2 S \longrightarrow S_{2} \quad \text { (given due to dimerisation) } \ \text { Initial } & \alpha & 0 \ \text { Final } & (1-\alpha) \quad \frac{\alpha}{2} \ \text { So, here } & i=\left(1-\frac{\alpha}{2}\right) \ & \Delta T_{bX}=\left(1-\frac{\alpha_{1}}{2}\right) K_{b(X)} \ \Delta T_{bY}= & \left(1-\frac{\alpha_{2}}{2}\right) K_{b(Y)} \end{array} $$
Given,
$$ \begin{aligned} \Delta T_{b(X)(s)} & =3 \Delta T_{b(Y)(s)} \ \left(1-\frac{\alpha_{1}}{2}\right) K_{b(X)} & =3 \times\left(1-\frac{\alpha_{2}}{2}\right) \times K_{b(Y)} \end{aligned} $$
or
$$ 2\left(1-\frac{\alpha_{1}}{2}\right)=3\left(1-\frac{\alpha_{2}}{2}\right) \quad\left[\because K_{b(X)}=2 K_{b(Y)}\right] $$
or
$$ \left.2\left(1-\frac{\alpha_{1}}{2}\right)=3\left(1-\frac{0.7}{2}\right) \quad \text { (as given, } \alpha_{2}=0.7\right) $$
or $\quad 4-2 \alpha_{1}=6-2.1$ or $2 \alpha_{1}=0.1$
$$ \text { so, } \quad \alpha_{1}=0.05 $$
