- Ideal
- Real
- Isotonic
- None of these
Answer: Option 1
For an ideal solution the heat of mixing or enthalpy change and volume change due to mixing is zero. Since volume change mixing $$ = {V^t}\left( {T,P} \right) - \sum {} {x_i}{V_i}\left( {T,P} \right).$$ Where $${V^t} = $$ total molar volume of the solution at temperature $$T$$, pressure $$P$$. $${V_i} = $$ molar volume of species when existed as pure species at same $$T$$, $$P$$ From summability : \[{{V}^{t}}=\sum{{}}{{x}_{i}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}\] where \[{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}=\] partial molar volume since for ideal solution \[\sum{{}}{{x}_{i}}{{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\smile}$}}{V}}}_{l}}=\sum{{}}{{x}_{i}}{{V}_{i}}\left( T,P \right)\] $$ \Rightarrow $$ Volume change of mixing $$= 0$$ Similarly, enthalpy change of mixing or heat of mixing $$= 0.$$