JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Solutions

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 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 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: 0.05

Solution

From the P-T curves, the elevation in boiling point for NaCl solutions:

For solvent X: $Δ T_{b} = 2$ K
For solvent Y: $Δ T_{b} = 1$ K

Since NaCl dissociates completely (

i = 2

) and the solutions are isomolal:

\frac{k _{b (X)}}{k _{b (Y)}} = \frac{2}{1} = 2

For dimerization:

2 S ⇌ S_{2}

Van't Hoff factor:

i = 1 - \frac{α}{2}

For solvent Y with

α = 0.7

i_{Y} = 1 - \frac{0.7}{2} = 0.65

Given that

Δ T_{b (X)} = 3 \times Δ T_{b (Y)}

k_{b (X)} \cdot m \cdot i_{X} = 3 \times k_{b (Y)} \cdot m \cdot i_{Y}

\frac{k _{b (X)}}{k _{b (Y)}} \cdot i_{X} = 3 \times i_{Y}

2 \cdot i_{X} = 3 \times 0.65

i_{X} = \frac{1.95}{2} = 0.975

For solvent X:

0.975 = 1 - \frac{α _{X}}{2}

\frac{α _{X}}{2} = 0.025

α_{X} = 0.05

Answer: 0.05

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer

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JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Solutions

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