At 35.0degrees, the vapour pressure of pure ethanol (C2H5OH) is 100.0mmHg and the vapour pressure of pure...

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At35.0degrees, the vapour pressure of pure ethanol (C2H5OH) is100.0mmHg and the vapour pressure of pure acetone (CH3COCH3) is360.0 mmHg
a) a solution is formed from 25.0g of ethanol and 15.0g ofacetone. Assume ideal behaviour and calculate the vapour pressureof each component above this solution.
b) calculate the mole fraction of acetone in the vapourpressure in equilibrium with the solution in part a).
c) Name a method that can be used to separate these twocomponents in the lab.

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3.9 Ratings (625 Votes)
Molecular wt of ethanol 46 So 250 g of ehanol 2546 0543 mole Molecular weight of acetone 58 So 15 g of acetone 1558 0259 mole total mol of    See Answer
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