1) At a pressure of 760 torr, formic acid (HCOOH; boiling point = 100.7 o C) and

Question

1) 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?

2) 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?

3) 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

1) let the mixture be 100 g

then

mass of HCOOH = 77.5

mass of water = 22.5

now

we know that

moles = mass / molar mass

so

moles of HCOOH = 77.5 / 46 = 1.6848

moles of H20 = 22.5 / 18 = 1.25

now

total moles = 1.6848 + 1.25 = 2.9348

now

mole fraction of HCOOH = moles of HCOOH / total moles

mole fraction of HCOOH = 1.6848 / 2.9348

mole fraction of HCOOH = 0.574

now

we know that

vapor pressure of HCOOH = mole fraction x vapor pressure of pure HCOOH

so

vapor pressure of HCOOH = 0.574 x 917 = 526.3

so

vapor pessure of HCOOH is 526.3 torr

2)

now

mole fraction of HCOOH + mole fraction of H20 = 1

so

0.574 + mole fraction of H20 =1

mole fraction of H20 = 0.426

now

vapor pressure of H20 = mole fraction x vapor pressure of pure H20

vapor pressure of H20 = 0.426 x 974

vapor pressure of H20 = 414.924

so

vapor pressure of H20 is 414.924 torr

3) now

total vapor presure = vapor pressure of HCOOH + vapor pressure of H20

= 526.3 + 414.924

= 941.2

so

the total vapor presure is 941.2 torr

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