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Two volatile substances, A and B are dissolved in one another and the resulting solution has a vapor pressure of 137 torr. If the mole fraction of B is 0.230 and the vapor pressure of pure B is 162 torr, what is the vapor pressure of pure A in torr?

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jufsanabriasa

Answer:

Vapor pressure of pure A is 129.5torr

Explanation:

When two or more liquids produce an ideal solution, the vapour pressure of the mixture follows the equation (Raoult's law):

[tex]P_T = P_aX_a + P_bX_b[/tex]

Where P is vapour pressure and X mole fraction of liquids A and B. Pt is the pressure of the solution

As mole fraction of B is 0.230, mole fraction of A is:

[tex]1 =X_a+X_b\\1=X_a+0.230\\X_a=0.77[/tex]

Computing in Raoult's law with the given values:

[tex]137 torr= P_a0.77 + 162torr*0.23\\137 torr = 0.77P_a+37.26torr\\99.74torr = 0.77P_a\\P_a=129.5torr[/tex]

Vapor pressure of pure A is 129.5torr

Eduard22sly

The vapor pressure of pure A in torr is 129.5 torr

How to determine the mole fraction of A

  • Mole fraction of B = 0.23
  • Mole fraction of solution = 1
  • Mole fraction of A =?

Mole fraction of A = 1 – 0.23

Mole fraction of A = 0.77

How to determine the vapor pressure of A

  • Total pressure (P) = 137 torr
  • Mole fraction of B (Xb) = 0.23
  • Vapor pressure of B (Pb) = 162 torr
  • Mole fraction of A (Xa) = 0.77
  • Vapour pressure of A (Pa) =?

P = XaPa + XbPb

137 = (0.77 × Pa) + (0.23 × 162)

137 = 0.77Pa + 37.26

Collect like terms

137 – 37.26 = 0.77Pa

99.74 = 0.77Pa

Divide both side by 0.77

Pa = 99.74 / 0.77

Pa = 129.5 torr

Thus, the vapor pressure of pure A is 129.5 torr

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