Consider the given data :
(a) $$HCl(g) + 10H_2O(l) \rightarrow HCl \cdot 10H_2O$$, $$\Delta H = -69.01$$ kJ mol$$^{-1}$$
(b) $$HCl(g) + 40H_2O(l) \rightarrow HCl \cdot 40H_2O$$, $$\Delta H = -72.79$$ kJ mol$$^{-1}$$
Choose the correct statement :

The enthalpy change recorded in an experiment always corresponds to the exact chemical equation that is written. For the same solute, if we write two different equations with two different amounts of solvent, we are dealing with two different processes and, therefore, two different enthalpy changes.

(a) $$HCl(g)+10H_2O(l)\rightarrow HCl\cdot10H_2O$$ has $$\Delta H=-69.01\ \text{kJ mol}^{-1}$$ (b) $$HCl(g)+40H_2O(l)\rightarrow HCl\cdot40H_2O$$ has $$\Delta H=-72.79\ \text{kJ mol}^{-1}$$

Observations:

1. Both enthalpy changes are negative, so dissolving gaseous HCl in water is an exothermic process, not an endothermic one. Hence Option A is wrong.

2. The two heats of solution are different because the final states of the system are different (one solution contains 10 molecules of water per HCl, the other 40). Therefore, the magnitude of the heat evolved depends on how much solvent is present. This directly supports Option B.

3. The heat of dilution is defined as the enthalpy change when an existing solution is diluted with more solvent. If the solution from (a) is further diluted to the composition of (b), the heat of dilution is $$\Delta H_{\text{dilution}}=\Delta H_{(b)}-\Delta H_{(a)}$$ $$\Delta H_{\text{dilution}}=-72.79\ \text{kJ mol}^{-1}-(-69.01\ \text{kJ mol}^{-1})$$ $$\Delta H_{\text{dilution}}=-3.78\ \text{kJ mol}^{-1}$$ The value is negative (exothermic). Option C states a positive 3.78 kJ mol$$^{-1}$$, so it is incorrect.

4. Because equations (a) and (b) describe two different final solutions, each has its own enthalpy of formation. They cannot both represent a single, unique heat of formation. Thus Option D is also incorrect.

Only Option B is correct: the heat of solution depends on the amount of solvent.