Question

The plot given shows $P-T$ curves (where $P$ is the pressure and $T$ is the temperature) for two solvents $X$ and $Y$ and isomolal solutions of $NaCl$ in these solvents. $NaCl$ completely dissociates in both the solvents.
On addition of an equal number of moles a non-volatile solute $S$ in equal amount (in kg) of these solvents, the elevation of the boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$, the degree of dimerization in solvent $X$ is:

Answer 1

From graph,

For solvent '$X$'; $\Delta T_{b\left(X\right)}=2$

$\Delta T_{b\left(X\right)}=i\times m_{NaCl}\times K_{b\left(X\right)}$ .......$\left(1\right)$

For solvent '$Y$'; $\Delta T_{b\left(Y\right)}=1$

$\Delta T_{b\left(Y\right)}=i\times m_{NaCl}\times K_{b\left(Y\right)}$ ........$\left(2\right)$

Equation $\left(1\right)/\left(2\right)$

$\Rightarrow \frac{K_{b\left(X\right)}}{K_{b\left(Y\right)}}=2$

For solute $S$

$2\left(S\right)\to S_2$

$1$ $-$

$1-\alpha$ $\alpha /2$

Now,

$i=(1-\alpha /2)$

$\Delta T_{b\left(X\right)\left(S\right)}=(1-\frac{{\alpha }_1}{2})K_{b\left(X\right)}m$

$\Delta T_{b\left(Y\right)\left(S\right)}=(1-\frac{{\alpha }_2}{2})K_{b\left(Y\right)}m$

Given $\Delta T_{b\left(X\right)\left(S\right)}=3\Delta T_{b\left(Y\right)\left(S\right)}$

$\left(1-\frac{{\alpha }_1}{2}\right)K_{b\left(X\right)}=3\times (1-\frac{{\alpha }_2}{2})\times K_{b\left(Y\right)}$

$2(1-\frac{{\alpha }_1}{2})=3(1-\frac{{\alpha }_2}{2})$ -------- 3

${\alpha }_2=0.7$

So, on substituting the value of ${\alpha }_2$ in 3 equation we get, ${\alpha }_1=0.05$.