Solutions and Colligative Properties - Result Question 27-1
27. The plot given below shows $p-T$ curves (where $p$ is the pressure and $T$ is the temperature) for two solvents $X$ and $Y$ and isomolal solution of $NaCl$ in these solvents. $NaCl$ completely dissociates in both the solvents.
On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in kg ) of these solvents, the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerisation in these solvents. If the degree of dimerisation is $0.7$ in solvent $Y$, the degree of dimerisation in solvent $X$ is ______.
(2018 Adv.)
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Answer:
Correct Answer: 27. $(0.05)$
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)} \quad$ ……(i)
Similarly for solution $y$
$ 1=i \times m_{\mathrm{NaCl}} \times K_{b(Y)} \quad$ ……(ii)
from Eq. (i) and (ii) above
$ \dfrac{K_{b(X)}}{K_{b(Y)}}=\dfrac{2}{1} \text { or } 2 \quad \text { or } \quad K_{b(X)}=2 K_{b(Y)} $
For solute $S$
$\text { So, here } i=(1-\dfrac{\alpha}{2})$
$ \Delta T _{b[X](s)}=\left(1-\dfrac{\alpha _1}{2}\right) K _{b(X)}$
$ \Delta T _{b[Y](s)}=\left(1-\dfrac{\alpha _2}{2}\right) K _{b(Y)}$
Given,
$ \begin{aligned} \Delta T_{b(X)(s)} & =3 \Delta T_{b(Y)(s)} \\ \left(1-\dfrac{\alpha_1}{2}\right) K_{b(X)} & =3 \times\left(1-\dfrac{\alpha_2}{2}\right) \times K_{b(Y)} \\ 2\left(1-\dfrac{\alpha_1}{2}\right) & =3\left(1-\dfrac{\alpha_2}{2}\right) \quad\left[\because K_{b(X)}=2 K_{b(Y)}\right] \\ 2\left(1-\dfrac{\alpha_1}{2}\right) & =3\left(1-\dfrac{0.7}{2}\right) \quad \text { (as given, } \alpha_2=0.7 \text { ) } \end{aligned} $
or $ 4-2 \alpha_1=6-2.1 \text { or } 2 \alpha_1=0.1 $
so, $ \alpha_1=0.05 $
BETA
