GATE 2024 MT – Question 56
An ideal solution is made by mixing 10 g of A and 50 g of B at 673 K. Its molar free energy of mixing is ______ kJ mol$^{-1}$ (one decimal place). Given: $R=8.314$ J mol$^{-1}$ K$^{-1}$; atomic weights are 40 g mol$^{-1}$ for A and 60 g mol$^{-1}$ for B.
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Correct answer: -3.2 to -2.8
Explanation
The amounts are $n_A=10/40=0.25$ mol and $n_B=50/60=5/6$ mol.
Hence $x_A=3/13$ and $x_B=10/13$.
For ideal mixing, $\Delta G_{mix}=RT(x_A\ln x_A+x_B\ln x_B)$.
Substitution gives about −3.02 kJ/mol.
The rounded molar value is **−3.0 kJ/mol**, rather than the total mixing free energy.