A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass =300 g mol^-1 ) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is \_\_\_\_ x 10^-2. (Nearest integer) [Given : K_ b of the solvent =5.0 K kg mol^-1 ] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Delta T_ b= i x K_ b x m 0.5= i x m x 5 i x m=(0.5 / 5)=0.1 i x a=(15 / 1000) (where a= moles of solute) Now, \ frac P_o- P_ S P^o= i X_ solute = i x frac a a+(150 / 300) \ = i x frac a1 / 2 =(15 / 1000 / 1 / 2) =(30 / 1000)=3 x 10^-2=3

A substance X 1 5 g dissolved in 150 g of a solvent Y m... | null (Chemistry)

JEE Mains 2026 21st January Shift 2

Chemistry

A substance ' X ' ( 1.5 g ) dissolved in 150 g of a solvent ' Y ' (molar mass

=300 \mathrm{~g} \mathrm{~mol}^{-1}

) led to an elevation of the boiling point by 0.5 K . The relative lowering in the vapour pressure of the solvent ' Y ' is

\_\_\_\_

\times 10^{-2}

. (Nearest integer)

\\

[Given :

\mathrm{K}_{\mathrm{b}}

of the solvent

=5.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ]\\

Assume the solution to be dilute and no association or dissociation of X takes place in solution.

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Correct Answer:

Explanation

\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}\\

0.5=\mathrm{i} \times \mathrm{m} \times 5\\

\mathrm{i} \times \mathrm{m}=\frac{0.5}{5}=0.1\\

\mathrm{i} \times \mathrm{a}=\frac{15}{1000}\\

(where

\mathrm{a}=

moles of solute) Now,

\\ \begin{aligned} & \frac{\mathrm{P}_{o}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^{o}}=\mathrm{i} \mathrm{X}_{\text {solute }}=\mathrm{i} \times \frac{\mathrm{a}}{\mathrm{a}+\frac{150}{300}} \\ & =\mathrm{i} \times \frac{\mathrm{a}}{1 / 2} =\frac{15 / 1000}{1 / 2} \\&=\frac{30}{1000}=3 \times 10^{-2}=3 \end{aligned}

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