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(C) For the formation of an ideal binary solution:$1$. The enthalpy of mixing is zero,i.e.,$\Delta H_{mix} = 0$.$2$. The entropy of mixing is positive

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(C) For the formation of an ideal binary solution:$1$. The enthalpy of mixing is zero,i.e.,$\Delta H_{mix} = 0$.$2$. The entropy of mixing is positive,i.e.,$\Delta S_{mix} > 0$,so $T \Delta S_{mix} > 0$.$3$. The Gibbs free energy of mixing is given by $\Delta G_{mix} = \Delta H_{mix} - T \Delta S_{mix}$. Since $\Delta H_{mix} = 0$,we have $\Delta G_{mix} = -T \Delta S_{mix}$. Because $\Delta S_{mix} > 0$,$\Delta G_{mix}$ must be negative $(\Delta G_{mix} Comparing these conditions with the given options:- $\Delta H_{mix}$ should be a straight line at $0$.- $T \Delta S_{mix}$ should be a positive curve (above the $0$ axis).- $\Delta G_{mix}$ should be a negative curve (below the $0$ axis).Option $C$ correctly shows $\Delta H_{mix} = 0$,$T \Delta S_{mix} > 0$,and $\Delta G_{mix} < 0$.

Which of the following graphs correctly represents the changes in thermodynamic properties during the formation of $1 \, mol$ of an ideal binary solution?

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