JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Solutions

Vessel-1 contains

w_{2} g

of a non-volatile solute

X

dissolved in

w_{1} g

of water. Vessel-2 contains

w_{2} g

of another non-volatile solute

Y

dissolved in

w_{1} g

of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van't Hoff factor for X is 1.2 times of that of

Y

for their respective concentrations.
The elevation of boiling point for solution in Vessel-1 is ____ % of the solution in Vessel-2.

Answer: 150

Solution

For Vessel-1:

(Δ T_{b})_{1} = i_{1} \times K_{b} \times \frac{w _{2} / GMM _{X}}{w _{1} /1000}

For Vessel-2:

(Δ T_{b})_{2} = i_{2} \times K_{b} \times \frac{w _{2} / GMM _{Y}}{w _{1} /1000}

Taking the ratio:

\frac{( Δ T _{b} ) _{1}}{( Δ T _{b} ) _{2}} = \frac{i _{1}}{i _{2}} \times \frac{GMM _{Y}}{GMM _{X}}

Given:

GMM_{X} = 0.8 \times GMM_{Y}

and

i_{1} = 1.2 \times i_{2}

\frac{( Δ T _{b} ) _{1}}{( Δ T _{b} ) _{2}} = \frac{1.2}{1} \times \frac{1}{0.8} = \frac{1.2}{0.8} = \frac{3}{2}

Percentage =

\frac{3}{2} \times 100 = 150%

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer

Practise more: all JEE Advanced PYQs on Solutions →

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Solutions

Solution

More JEE Advanced questions on Solutions