Raoult s law?

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2017-10-02 7:52 pm

Solution B was prepared by mixing 2.23 g of liquid propanone with 3.58 g of liquid butanone at 30°C in a closed container.

The vapour pressure of pure propanone (C3H6O, molar mass = 58.1 g/mol) at 30°C is 285 mm Hg and that of pure butanone (C4H8O, molar mass = 72.1 g/mol) is 121 mm Hg at the same temperature. From this information, and assuming Raoult s Law applies to his mixture, calculate the values below at 30°C.

(a) the mole fraction of propanone in Solution B.
(b) the mole fraction of butanone in the solution.
(c) the partial pressure (in mm Hg) of propanone above Solution B.
(d) the partial pressure (in mm Hg) of butanone above Solution B.
(e) the mole fraction of propanone in the vapour phase.

回答 (1)

wanszeto

2017-10-02 8:33 pm

(a)
No. of moles of C₃H₆O = (2.23 g) / (58.1 g/mol) = 0.0384 mol
No. of moles of butanone = (3.58 g) / (72.1 g/mol) = 0.0497 mol

Mole fraction of C₃H₆O, X(C₃H₆O) = 0.0384 / (0.0384 + 0.0497) = 0.436

(b)
Mole fraction of C₄H₈O, X(C₄H₈O) = 0.0497 / (0.0384 + 0.0497) = 0.564

(c)
Assume that the solution is ideal.
Partial pressure of C₃H₆O, P(C₃H₆O) = X(C₃H₆O) * P°(C₃H₆O) = 0.436 * (285 mmHg) = 124 mmHg

(d)
Partial pressure of C₄H₈O. P(C₄H₈O) = X(C₄H₈O) * P°(C₄H₈O) = 0.564 * (121 mmHg) = 68.2 mmHg

(e)
At constant T and V, the ratio in moles of gases is equal to the ratio in there partial pressure.
Mole fraction in C₃H₆O in the vapour phase = 124 / (124 + 68.2) = 0.645

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