Methanol (CH3OH) – ethanol (CH3CH2OH) solutions are almost
ideal. If the vapour pressures of methanol and ethanol at 335K are
(8.1 x 10*4) and (4.5 x 10*4)Nm -2 respectively, calculate the
composition (by mole fraction) of the vapour over a mixture of 64g
of methanol and 46 g of ethanol at 335K
(Relative atomic masses: H:=1.0, C=12, O=6)
Vapour mixture calculation
2009-01-02 11:16 pm
回答 (1)
2009-01-03 5:59 pm
✔ 最佳答案
PM: Partial pressure of methanol (CH3OH) in vapour mixture
XM: Mole fraction of methanol (CH3OH) in liquid mixture
PMo = vapour pressure of methanol (CH3OH) = 8.1 * 104 N m-2
PE: Partial pressure of ethanol (CH3CH2OH) in vapour mixture
XE: Mole fraction of ethanol (CH3OH) in liquid mixture
PEo = vapour pressure of ethanol (CH3CH2OH) = 4.5 * 104 N m-2
Molar mass of CH3OH = 12 + 1*4 + 16 = 32 g mol-1
Molar mass of CH3CH2OH = 12*2 + 1*6 + 16 = 46 g mol-1
No. of moles of CH3OH used = 64/32 = 2 mol
No. of moles of CH3CH2OH = 46/46 = 1 mol
In vapour:
PM = XM*PMo = (2/3) * (8.1 * 104) = 5.4 * 104 N m-2
PE = XE*PEo = (1/3) * (4.5 * 104) = 1.5 * 104 N m-2
Total pressure P = 5.4 * 104 + 1.5 * 104 = 6.9 * 104 N m-2
Mole fraction of CH3OH in vapour = (5.4 * 104)/( 6.9 * 104) = 18/23 (or 0.783)
Mole fraction of CH3CH2OH in vapour = (1.5 * 104)/( 6.9 * 104) = 5/23 (or 0.217)
=
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