The vapor pressure of ethanol (C2H5OH) at 20 degree C is 44 mmHg, and the vapor
ID: 870096 • Letter: T
Question
The vapor pressure of ethanol (C2H5OH) at 20 degree C is 44 mmHg, and the vapor pressure of methanol (CH3 OH) at the same temperature is 94 mmHg. A mixture of 27.5 g of methanol and 44.5 g of ethanol is prepared and can be assumed to behave as an ideal solution. Calculate the vapor pressure of methanol and ethanol above this solution at 20 degree C. Be sure to report your answers to the correct number of significant figures. Calculate the mole fraction of methanol and ethanol in the vapor above this solution at 20 degree C.Explanation / Answer
Let us denote ethanol by the subscript A and methanol by the subscript B
Given :
Pure vapor pressure of ethanol, PAo = 44 mm Hg
Pure vapor pressure of methanol, PBo = 94 mm Hg
Moles of ethanol taken, NA = Mass taken / MW = 44.5/46 = 0.967 moles
Moles of methanol taken, NB = Mass taken / MW = 27.5/32 = 0.859 moles
Mole fraction of ethanol, XA = NA/(NA + NB) = 0.53
Mole fraction of methanol, XB = 1 - XA = 0.47
Using Raoult's law of partial pressure,
Vapor pressure of ethanol above solution = PA = PAo*XA = 44*0.53 = 23.32 mm Hg
Vapor pressure of methanol above solution = PB = PBo*XB = 94*0.47 = 44.18 mm Hg
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