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AL CHEM~

用戶92111

2009-04-13 5:43 pm

In an experiment, 2.4 g of ethanoic acid undergoes complete
vaporisation in a closed containerof volume 1.5 dm^3 maintained at
a constant temperature of 450 K. The vapour produced exerts a
pressure of 69.8 kPa.

a) Assuming that the vapour of ethanoic acid behaves as an ideal gas ,
calculate the apparent molar mass of ethanoic acid under the
experimental conditions .

b)Comment on the value obtained in (i) with respect to the molecular
formula of ethanoic acid .

回答 (1)

老爺子

2009-04-13 7:45 pm

✔ 最佳答案

a)
m = 2.4 g
P = 69.8 x 103 Pa
V = 1.5 dm3 = 1.5 x 10-3 m3
T = 450 K
R = 8.31 J mol-1 K-1

PV = nRT
PV = (m/M)RT
M = mRT/PV

Molar mass of CH3COOH, M
= mRT/PV
= (2.4 x 8.31 x 450)/[(6.98 x 103) x (1.5 x 10-3)]
= 85.7 g mol-1

b)
Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1

The actual molar mass is greater than the theoretical molar mass of CH3COOH.
It is because a portion of CH3COOH molecules exist as dimers by linking two molecules by hydrogen bonds, and another portion of CH3COOH molecules exist as monomer. The structure of a CH3COOH is shown below:

圖片參考：http://members.optushome.com.au/scottsoftb/AceticDimer.gif

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