At ?10 Degree C, the vapor pressure of ethyl bromide is 90.0 Torr and that of et

Question

At ?10 Degree C, the vapor pressure of ethyl bromide is 90.0 Torr and that of ethyl chloride is 302.5 Torr. Assume that the solution ?s ?deal. Assume there is only a trace of liquid present and the mole fraction of ethyl chloride in the vapor is 0.82 and answer these questions: a. What is the total pressure and the mole fraction of ethyl chloride in the liquid? b. If there are 6.50 mol of liquid and 3.80 mol of vapor present at the same pressure as in part (a), what is the overall composition of the system?

Explanation / Answer

According to Raoult's law:

pT = pe + pm = peoXe + pmoXm

Where:

pe & pm are partial vapor pressures of ethylbromide and ethylchloride, respectively.

peo & pmo are pure vapor pressures of ethylbromide and ethylchloride, respectively.

Xe & Xm are the mole fractions of ethylbromide and ethylchloride in the liquid phase, respectively.

However, according to Dalton's law:

pT = pe + pm = pTYe + pTYm

Where:

Ye & Ym are the mole fractions of ethylbromide and ethylchloride in the vapour phase, respectively.

Using above equations

Given peo & pmo are 90 and 302.5 torr

Ym = 0.82 and Ye = 0.18

peoXe = pTYe   ====>90*Xe = pT*0.18

===>pT = 500Xe

pmoXm = pTYm  =====> 302.5 *Xm = pT *0.82

===>pT= 368.90Xm

Xe + Xm =1 so Xm = 1-Xe

368.90[1-Xe] = 500 Xe

Xe = 0.424 and Xm = 0.576

A] pT = 0.576*368.9 = 212.48

Xm = 0.576

B] 6.5 mol of liquid and 3.8 mol of vapor

moles of ethyl chlroide in liquid= 0.576 * 6.5 = 3.744

moles of ethyl bromide in liquid = 2.756

moles of ethylchloride in vapor = 0.82*3.8 = 3.116

moles of ethyl bromide in vapor = 0.684

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