At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 oC) and wat

Question

At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 oC) and water (H2O; boiling point = 100.0 oC) form an azeotropic mixture, boiling at 107.1 oC, that is 77.5% by mass formic acid. At the boiling point of the azeotrope (107.1 oC), the vapor pressure of pure formic acid is 917 torr, and that of pure water 974 torr. If the solution obeyed Raoult's law for both components, what would be the vapor pressure (in torr) of formic acid at 107.1 oC? If the solution obeyed Raoult's law for both components, what would be the vapor pressure of the water (in torr) at 107.1 oC? If the solution obeyed Raoult's law for both components, what would be the total vapor pressure (in torr) at 107.1 oC?

Explanation / Answer

The mol wt of HCOOH = 46; H2O =18. If the wt% composition of HCOOH in the azeotrope is 77.5% then you can write assuming that the mol fracn of acid is xA and that of water xB--->> (46*xA)/(46*xA +18*xB) = 0.775, solving which you obtain xA = 0.57 and xB = 0.43. If the azeotrope obeyed Raoult's Law, P =P°A*xA + P°BxB where P°A and P°B are the vapor pressures of pure HCOOH and H2O at 107.1 °C respectively. So substituting the value of xA and xB in the formula yields 949.49. Instead of this, the value in reality is only 760 torr, because this azeotrope shows a negative deviation from Raoult's Law.

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