At 63.5oC the vapor pressure of water is 175 torr and that of ethanol (C2H5OH) i

Question

At 63.5oC the vapor pressure of water is 175 torr and that of ethanol (C2H5OH) is 400 torr. A solution is
made by mixing equal masses of water and ethanol.
a. What is the molality of ethanol in the solution?
b. What is the mole fraction of ethanol in the solution?
c. Assuming ideal solution behavior, what is the vapor pressure of the solution at 63.5oC?
d. What is the mole fraction of ethanol in the vapor above the solution?

I did parts B, C and D and I believe I got it right but the one I have problem is with part A. '/

Explanation / Answer

sol Assume X grams of H2O and C2H5OH are mixed. The molar masses of H2O and C2H5OH are 18.015 g/mol and 46.068 g/mol, respectively. (a) Thus the mole numbers of H2O and C2H5OH in the mixture are X/18.015 and X/46.068, respectively. The mole fraction of ethanol in the solution is: (X/46.068)/(X/18.015 + X/46.068) = 18.015/(18.015+46.068) = 0.28112 (b) Of course the mole fraction of water in the solution is: 1-0.28112 = 0.71888 The total vapor pressure of the solution at 63.5ºC is: 0.28112*400+0.71888*175 = 238 (Torr) (c) 0.28112*400/238 = 0.472 ans

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