Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 4 April, Evening Shift — Question 12

Consider the given data:

(a) HCl(g)+10H2O(I)→HCl.10H2OΔH=−69.01\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_{2} \mathrm{O} \Delta \mathrm{H}=-69.01 kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}

(b) HCl(g)+40H2O(I)→HCl.40H2OΔH=−72.79\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_{2} \mathrm{O} \Delta \mathrm{H}=-72.79 kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}

Choose the correct statement:

  1. Option A:

    Dissolution of gas in water is an endothermic process.

  2. Option B:

    The heat of solution depends on the amount of solvent.

  3. Option C:

    The heat of formation of HCl solution is represented by both (a) and (b).

  4. Option D:

    The heat of dilution for the HCl(HCl.10H2O\mathrm{HCl}\left(\mathrm{HCl} .10 \mathrm{H}_{2} \mathrm{O}\right. to HCl.40H2O\mathrm{HCl} .40 \mathrm{H}_{2} \mathrm{O} ) is 3.78 kJ mol−13.78 \mathrm{~kJ} \mathrm{~mol}^{-1}.

    Correct

Answer: D

Step-by-step solution

HCl(g)+∞H2O(I)→HCl(aq)\mathrm{HCl}(\mathrm{g})+\infty \mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{HCl}(\mathrm{aq})

Heat released the above process is heat of solution.

From reaction (b) - (a) we get heat of dilution for HCl(HCl.10H2O\mathrm{HCl}\left(\mathrm{HCl} .10 \mathrm{H}_{2} \mathrm{O}\right. to HCl.40H2O)\left.\mathrm{HCl} .40 \mathrm{H}_{2} \mathrm{O}\right) as 3.78 kJ mol−13.78 \mathrm{~kJ} \mathrm{~mol}^{-1}.

Answer key and solution verified before publishing.

Practise Thermodynamics & Thermochemistry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2025
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Thermochemistry and Enthalpy Changes
Consider the given data: (a) HCl ( g )+10 H 2 O ( I ) rightarrow HCl… | JEE Main 2025 PYQ with Solution · DhiX AI