
Concept explainers
A mixture of liquids A and B exhibits ideal behavior. At 84°C, the total vapor pressure of a solution containing 1.2 moles of A and 2.3 moles of B is 331 mmHg. Upon the addition of another mole of B to the solution, the vapor pressure increases to 347 mmHg. Calculate the vapor pressures of pure A and B at 84°C.

Interpretation:
Vapor pressure of pure A and B solutions at
Concept introduction:
Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.
Where,
Answer to Problem 12.133QP
Vapor pressure of pure A solution =
Vapor pressure of pure B solution =
Record the given data
Vapor pressure of the solution =
Amount of solution A =
Amount of solution B =
Increased vapor pressure of the solution =
Explanation of Solution
Given data are recorded as shown.
To calculate mole fraction of total solution
Substituting in
By plugging in the values of mole fraction of solution A and B, mole fraction of total solution has calculated.
To calculate mole fraction of total solution after additional moles of B
Substituting in
By plugging in the values of mole fraction of solution A and mole fraction of solution after addition of solution B, after additional moles of B has calculated.
To calculate vapor pressure of pure solution of A and B
Substituting equation (1) into (2) we get,
Substituting the value of
By substituting the equation (1) into equation (2), the vapor pressure of pure solution B has calculated and by subtracting this value into
Vapor pressure of pure solution A was calculated as
Vapor pressure of pure solution B was calculated as
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Chapter 12 Solutions
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