1. Ideal
  2. Real
  3. Isotonic
  4. None of these

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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 $$= 0.$$

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