Chemistry - Thermodynamics - I Question with Solution | TestHub

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