Question

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

Answer 1

From the graph we can note,
$\Delta T_b$ for solution $X$ i.e.,
$\Delta T_{b\left(X\right)}=362-360=2$

Likewise, $\Delta T_b$ for solution $Y$ i.e., $\Delta T_{b\left(Y\right)}=368-367=1$

Now by using the formula,
$\Delta T_b=i\times \text{molality of solution}\times K_b$

For solvent $X$,
$2=i\times m_{NaCl}\times K_{b\left(X\right)}$

Similarly for solvent $Y$,
$1=i\times m_{NaCl}\times K_{b\left(Y\right)}$ ...(ii)

$\therefore \frac{K_{b\left(X\right)}}{K_{b\left(Y\right)}}=\frac{2}{1}=2or{K}_{b\left(X\right)}=2K_{b\left(Y\right)}$

For solute $S$,

$2S\to S_2$

Initial $\alpha$ 0

Final $(1-\alpha )\frac{\alpha }{2}$

So, $i=(1-\frac{\alpha }{2})$

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

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

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

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

$2(1-\frac{{\alpha }_1}{2})=3(1-\frac{a_2}{2})[\because K_{b\left(X\right)}=2K_{b\left(Y\right)}]$

$2(1-\frac{{\alpha }_1}{2})=3(1-\frac{0.7}{2})(\text{as given,}{\alpha }_2=0.7)$

$4-2{\alpha }_1=6-2.12{\alpha }_1=0.1\therefore {\alpha }_1=0.05$