Question

Q1: A mixture of benzene water is boiling at atmospheric pressure . A benzene is immecible in water. Determine the boiling boint of mixture and the composition of the vapor. 4points If the benzene amount is 10 mol. How much steam you need to vaporize it. 3points Vapor pressure data of pure components are as follows: Temperature water K Pua (mm Hg) °C Pbenzene (mm Hg) 308.5 325.9 345.8 353.3 35.3 52.7 72.6 80.1 43 106 261 356 150 300 600 760

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Abdul-Rahim Taysir

Answer 1

A)

Benzene - A and water - B

are immiscible

So

At boiling point

P = P_as + P_bs

P_as - vapor pressure of benzene

P_bs - vapor pressure of water

P = 1 atm = 760 mmHg

The boiling point ofod mixture will be less than boiling point of either of two pure components

The vapor pressures are given in table above

Vapor pressures are calculated by interpolation

Temperature	V.P(mmHg) (water)	V.P(mmHg) (benzene)
35.3	43	150
52.7	106	300
72.6	261	600
80.1	356	760

The sum of both vapor pressures should be equal to 760 mmHg

760 = P_as+ P_bs

At 52.7°C

Pas =300 mmHg

Pbs = 106 mmHg

Pas + Pbs = 406 mmHg which is lesser than 760 mmHg

At 72.6°C

Pas = 261 mmHg

Pbs = 600 mmHg

Pas + Pbs = 861 mmHg which is greater than 760 mmHg

This shows that booking point of mixture lies between 52.7°C to 72.6°C

At 52.7°C - sum of vapor presssures = 406 mmHg

At 72.6°C - sum of vapor presssures = 861 mmHg

By interpolation ,

At Pas + Pbs = 760 mmHg

Boiling temperature of mixture = 68.182 °C

At 68.182°C

P_as = 533.4065 mmHg

P_bs = 226.593 mmHg

Applying raoults law

yP = P_as

y (760) = 533.4065

y = 0.70185

Benzene - 0.70185

(1-y) = 0.29815

Water - 0.29815

Both are immiscible

B) . Amount of benzene to be vaporized = 10 mol

For steam distillation

$\frac{W_{A}}{W_{b}}= \frac{P_{as}}{P_{bs}}\left ( \frac{M_{A}}{M_{B}} \right )$

Benzene to be evaporated = 10 mol

W_A/M_A = 10 mol

M. W of water = 18 g /mol

M. W of benzene = 78 g/mol

$\frac{W_{A}(M_{B})}{W_{b}(M_{A})}= \frac{P_{as}}{P_{bs}}$

$\frac{10(M_{B})}{W_{b}}= \frac{533.4065}{226.593}$

M_B/W_{B =} 4. 24803 moles

Moles of steam requires to evaporate 10 mol benzene

Mass of steam required = 4.24804(18) =

76.464 g

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