Consider the given data :
(a) HCl(g)+10 H₂ O(l) → HCl .10 H₂ O Δ H=-69.01 kJ mol⁻¹
(b) HCl(g)+40 H₂ O(l) → HCl .40 H₂ O Δ H=-72.79 kJ mol⁻¹
Choose the correct statement :

Solution

Chemistry ma’am · 4 steps

Same HCl, different water, different Δ H: the heat of solution depends on how much solvent you use.

1. Read the data. Both reactions dissolve 1 mol of HCl(g). Only the amount of water changes, from 10 mol to 40 mol, and the enthalpy changes from -69.01 to -72.79 kJ mol⁻¹. So the heat of solution depends on the amount of solvent.
2. Heat of dilution. Going from HCl.10H₂O to HCl.40H₂O is the second reaction minus the first.

   Δ H_(dil) = (-72.79) - (-69.01) = -3.78 kJ mol⁻¹

3. Check the rest. The dilution heat is negative, so heat is released, not +3.78. Both Δ H values are negative, so dissolving this gas is exothermic. Both reactions start from HCl(g), not from the elements H₂ and Cl₂, so they are heats of solution, not heats of formation.
4. So. The correct statement is (c): the heat of solution depends on the amount of solvent.

The idea to keep. Integral heat of solution changes with the amount of solvent. Subtract two such values to get the heat of dilution, and always keep its sign.

Answer from the NTA final answer key. Solution by PrepWiser.