Methanol and ethanol form a nearly ideal solution at 300K.A solution is made by mixing 32 g of methanol and 23 g of ethanol at 300 K .Calculate the partial pressure of its constituents and the total pressure of the solution.( At 300 K p0 CH3OH = 90 mm Hg p0 C2H5OH = 51 mm Hg

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Moles of methanol=MassMolar mass=32 g32 gmol-1=1 molMoles of ethanol=MassMolar mass=23 g46 gmol-1=0.5 molMole fraction of methanol, XCH3OH=Moles of methanolTotal molesor, XCH3OH=11+0.5=11.5=0.67Mole fraction of ethanol,XC2H5OH=1-XCH3OH =1-0.67=0.33Applying Raoult's law, Partial Vapour pressure of methanol = 0.67×90=60.3 mmPartial Vapour pressure of ethanol = 0.33×51=16.83 mmTotal pressure= 60.3 mm+16.82 mm=77.12 mm

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