A solution is made of 398 g toluene (C 7H 8, 92.1 g/mol) and 734 g benzene (C 6H 6, 78.1 g/mol). At 20.0 o C, the vapor pressure of pure toluene is 22 torr and the vapor pressure of pure benzene is 75 torr. What is the total pressure of toluene plus benzene above the solution at 20.0 ºC?

58.305torr

Explanation:

We shall solve this using Raoult law.

Mathematical expression is;

Ptotal=XaPa + XbPb

Where Xa, Xb and Pa Pb are the mole fraction and vapor pressure of each component. we shall now solve this by steps

STEP1: Find the mole of benzene and Toluene;

Mole= mass÷molar mass

For toluene:

Mass= 398g

Molar mass= 92.1g/mol

Mole= 398÷92.1 =4.32mol

For benzene:

Mass=734g

Molar mass= 78.1g/mol

Mole= 734÷78.1 = 9.398mol

STEP2: Find the mole fraction of each component:

Let Xb be mole fraction for benzene

Let Xt be mole fraction for toluene.

Therefore; nB+nT = X

nB and nT are the mole of component B and T

X= 4.32+9.398= 13.718

Mole fraction for Toluene;

Xt= nT÷X

4.32÷13.718= 0.315

Mole fraction for benzene;

Xb= nB÷X

9.398÷13.718= 0.685

STEP3: Find the total total pressure;

We will use Rauolt law.

Ptotal= XbPb + XtPt

Pt = 22torr

Pb = 75torr

Therefore Ptotal is;

(0.315×22) + (0.685×75) = 6.93+51.375= 58.305

Therefore the total pressure is 58.305torr

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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