Question

The plot given below shows P - T curves (where P is 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 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 than 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

As we know,

$2=2\times K_{b\left(x\right)}m$
$1=2K_{b\left(y\right)}m$
$\frac{K_{b\left(x\right)}}{K_{b\left(y\right)}}=2$
$\Delta T_{b\left(x\right)}=(1-\frac{{\alpha }_1}{2})K_{b\left(x\right)}m\underset{1-\alpha }{2S}\rightleftharpoons \underset{\frac{\alpha }{2}}{S_2}$
$\Delta T_{b\left(y\right)}=(1-\frac{{\alpha }_1}{2})K_{b\left(y\right)}mi=1-\alpha +\frac{\alpha }{2}$
$3=\frac{\Delta T_{b\left(x\right)}}{\Delta T_{b\left(y\right)}}=\frac{(1-\frac{{\alpha }_1}{2})K_{b\left(x\right)}}{(1-\frac{0.7}{2})K_{b\left(y\right)}}i=1-\frac{\alpha }{2}$
$3=\frac{(1-\frac{{\alpha }_1}{2})\times 2}{(1-\frac{0.7}{2})}$
$(1-\frac{{\alpha }_1}{2})=\frac{3\times 0.65}{2}=1.5\times 0.65$
${\alpha }_1=0.05$

Hence, the degree of dimerization in solvent X is 0.05.