Explain the enthalpy of dilution.

Enthalpy of dilution is the enthalpy change observed when a specified amount of solvent is added to a solution at a constant temperature and pressure.
For example,the enthalpy change for dissolving $1 \ mol$ of gaseous hydrogen chloride in $10 \ mol$ of water is represented by:
$HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O ; \Delta H = -69.01 \ kJ/mol$
Consider the following enthalpy changes for different amounts of water:
$(S-1) \ HCl_{(g)} + 25 \ H_2O_{(l)} \rightarrow HCl \cdot 25 \ H_2O ; \Delta H = -72.03 \ kJ/mol$
$(S-2) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O ; \Delta H = -72.79 \ kJ/mol$
$(S-3) \ HCl_{(g)} + \infty \ H_2O_{(l)} \rightarrow HCl \cdot \infty \ H_2O ; \Delta H = -74.85 \ kJ/mol$
As more solvent is added,the enthalpy of solution approaches a limiting value at infinite dilution,as shown in $(S-3)$.
To find the enthalpy of dilution between two concentrations,subtract $(S-1)$ from $(S-2)$:
$(HCl \cdot 25 \ H_2O) + 15 \ H_2O \rightarrow HCl \cdot 40 \ H_2O ; \Delta H = [-72.79 - (-72.03)] = -0.76 \ kJ/mol$
This value,$-0.76 \ kJ/mol$,is the enthalpy of dilution. It depends on the initial concentration of the solution and the amount of solvent added.