Consider the given data :

(a) $\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(b) $\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O} \Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Choose the correct statement :

<p>From the given information</p> <p>$\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.</p> <p>$$\mathrm{HCl}(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}\quad\text{..... (1)}$$</p> <p>$$\begin{aligned} & \Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}} \\ & \mathrm{HCl}(\mathrm{~g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}\quad\text{..... (2)} \end{aligned}$$</p> <p>$\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$</p> <p>Hence heat of solution depends upon amount of solvent</p> <p>By equation ............. (2) $-$ equation ........... (1)</p> <p>$$\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$$</p> <p>So Heat of dilution $=-72.79-(-69.01)$</p> <p>$=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$</p> <p>Hence option (3) is incorrect.</p> <p>For heat of formation reactant should be in elemental form hence option (4) is incorrect</p>