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almond37 [142]
3 years ago
9

At 25°C the vapour pressure of pure pentane is 511 torr and that of hexane is 150. torr.

Chemistry
1 answer:
Genrish500 [490]3 years ago
3 0

Based on the vapor pressure of the pure substances, the mole fraction of pentane is 0.45 and mole fraction of hexane at equilibrium is 0.60.

<h3>What is vapor pressure of a liquid?</h3>

Vapor pressure of of a liquid is the pressure that is excreted on the walls of a sealed vessel by the molecules of the liquid that have been converted to gaseous phase.

TO determine the mole fraction of pentane in the solution, we use the vapor pressure of the pure substances.

Let the mole fraction of pentane in the mixture be X.

The mole fraction of hexane will be 1 - X

  • The vapor pressure of the solution = (mole fraction of pentane × vapor pressure of pentane) + (mole fraction of hexane × vapor pressure of hexane)

Vapor pressure of solution = 383 torr

(mole fraction of pentane × vapor pressure of pentane) + (mole fraction of hexane × vapor pressure of hexane) = 383 torr

X × 511 + (1 - X) × 150 = 383

511X - 150X + 150 = 383

511X = 233

X = 0.45

Also , at equilibrium, the vapor pressure of pentane = 0.45 × 511 = 230 torr

mole fraction of hexane in the vapor that is in equilibrium with this solution = 230/383

mole fraction of hexane at equilibrium = 0.60

Therefore, the mole fraction of pentane is 0.45 and mole fraction of hexane at equilibrium is 0.60.

Learn more about vapor pressure at: brainly.com/question/26006166

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Explanation:

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There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

nn_{HBr}=100 mol HBr

n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78

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