Assume that 1-bromobutane and 1-chlorobutane form an ideal solution. At 273 P* c

Question

Assume that 1-bromobutane and 1-chlorobutane form an ideal solution. At 273 P* chloro = 3790. Pa and p* bromo = 1394 Pa. When only a trace of liquid is present at 273 K, X^(g) chloro = 0.63 a. Calculate the total pressure above the solution. b. Calculate the mole fraction of 1-chlorobutane in the solution. c. What value would x^(tot)chloro have in order for there to be 4.05 mol of liquid and 3.65 mol of gas at a total pressure equal to that in part (a)? [Note: This composition is different from that of part (a)].

Explanation / Answer

According to Raoult's law:

pT = pe + pm = peoXe + pmoXm

Where:

pe & pm are partial vapor pressures of 1chloro butane and 1Bromo butane, respectively.

peo & pmo are pure vapor pressures of 11chloro butane and 1Bromo butane, respectively.

Xe & Xm are the mole fractions of 11chloro butane and 1Bromo butane in the liquid phase, respectively.

However, according to Dalton's law:

pT = pe + pm = pTYe + pTYm

Where:

Ye & Ym are the mole fractions of 11chloro butane and 1Bromo butane in the vapour phase, respectively.

Using above equations

Y chloro (g)= 0.63

Y bromo (g)= 0.37

Comparing two results we can say

peoXe = pTYe

3790*Xe = Pt *0.63 ======> Pt = 6015.87 Xe

pmoXm = pTYm

1394 * Xm = pT *0.37 =====>Pt = 3767.56Xm

we Xe + Xm = 1 ====> Xm =1-Xe

therefore   6015.87 Xe = [1-Xe]3767.56

b] Xe = 0.385 and Xm =0.615

a]Pt = 3767.56 *0.615 = 2317.04

c] X chloro [total] = X in solution + Y in gas

=[ 0.385*4.05 + 0.63 *3.65]/ 4.05+3.65 = 3.85 /7.7 = 0.5

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