(a) A solution composed of 55.5 g of carbon disulfide and 104.3 g of dimethoxyme
ID: 988046 • Letter: #
Question
(a) A solution composed of 55.5 g of carbon disulfide and 104.3 g of dimethoxymethane is made. Assuming an ideal solution is formed, what is the vapor pressure of this solution?
(b) What is the composition (in terms of mole fraction) in the vapor phase?
(c) Solutions of carbon disulfide and dimethoxymethane are in fact not ideal - an experimental measurement results in a vapor pressure of 690.3 torr for the solution composition of part (a). Explain how this deviates from Raoult’s law, and explain the physical origin (cause) of this deviation.
Explanation / Answer
a)
moles of carbon disulfide CS2 = 55.5 / 76 = 0.73
moles of dimethoxymethane = 104.3 / 76 = 1.37
total moles = 1.37 + 0.73 = 2.1
mole fraction of carbon disulfide = XC = 0.73 / 2.1 = 0.35
mole fraction of dimethoxymethane = XM = 1.37 / 2.1 = 0.65
vapour pressure of total mixture = XC . VC + XM . VM
= 0.35 x 514.5 + 0.65 x 587
= 561.6 torr
vapour pressure of solution = 561.6 torr
b)
mole fraction of carbon disulfide = XC = 0.73 / 2.1 = 0.35
mole fraction of dimethoxymethane = XM = 1.37 / 2.1 = 0.65
total pressure = 561.6 torr
Yc = mole fraction in vapor phase
PC XC = Ptotal . YC
514.5 x 0.35 = 561.6 x YC
YC = 0.321
mole fraction of CS2 = 0.321
PM XM = Ptotal . YM
587 x 0.65 = 561.6 x YM
YM = 0.679
mole fraction of dimethoxymethane = 0.679
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