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Q.

Two moles of an ideal solution is prepared by mixing pure liquids ‘A’ and ‘B’ at 27ºC. The mole percent of ‘A’ in the vapor above the solution is 60. Vapor pressure of pure liquids A & B are PA0=120mm Hg and PB0=80mm Hg respectively. What is the magnitude of free energy change of mixing ΔmixGfor solution in calories? (Take, R=2 cal K-1 mol-1,log2=0.3,log3=0.48,log5=0.7,2.303≃2.3 )

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answer is 828.

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Detailed Solution

ΔmixG=ΔmixH−TΔmixSFor ideal solution, ΔmixH=0ΔmixG=−TΔmixSΔmix=−2.303R×ntotal∑xilogxi≃−2.3×R×ntotal∑xilogxi1Ps=0.6120+0.480⇒Ps=100100=120×XAL+80×1−XA1XAl=12XBl=12ΔmixS=−2.3×2×212log12+12log12cal=2.76calΔmixG=−300×2.76cal=−828calΔmixG=828cal
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