A solution is prepared by mixing 2.88 mol of acetone with 1.45 mol of cyclohexane at 30°C. 2nd attempt Calculate the χacetone and χcyclohexane in the vapor above the solution. P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr.

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21-01-2018
Chemistry

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A solution is prepared by mixing 2.88 mol of acetone with 1.45 mol of cyclohexane at 30°C. 2nd attempt Calculate the χacetone and χcyclohexane in the vapor above the solution. P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr.

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Dejanras Dejanras · Answer 1 · 2018-01-30T21:31:40+01:00

First calculate the mole fraction of each substance:
acetone: 2,88 mol ÷ (2,88 mol + 1,45 mol) = 0,665.
cyclohexane: 1,45 ÷ (2,88 mol + 1,45 mol) = 0,335.
Raoult's Law:
P(total) = P(acetone) · χ(acetone) + P(cyclohexane) · χ(cyclohexane).
P(total) = 229,5 torr · 0,665 + 97,6 torr · 0,335.
P(total) = 185,3 torr.
χ for acetone: 229,5 torr · 0,665 ÷ 185,3 torr = 0,823.
χ for cyclohexane: 97,6 torr · 0,335 ÷ 185,3 torr = 0,177.