Question

A sample of a gas mixture contains the following quantities of three gases.

compound	mass
CO	2.68 g
CO2	3.69 g
SF6	2.13 g

The sample has:

volume = 2.50 L
temperature = 16.6 °C

What is the partial pressure for each gas, in mmHg?
What is the total pressure in the flask?

CO ___ mmhg

CO₂ __mmhg

SF_{6 ____} mmHg

total ___ mmHg

Answer 1

no of mole of CO = w/mwt = 2.68/28 = 0.096 mole

no of mole of CO2 = 3.69/44 = 0.084 mole

no of mole of SF6 = 2.13/146.06 = 0.146 mole

total no of mole = 0.096+0.084+0.146 = 0.326 mole

Total pressure(Ptotal) = nRT/V

           = 0.326*0.0821*(16.6+273.15)/2.5

            = 3.1 atm

            = 2356 mm of Hg

molefraction of CO = nCO/nTOtal = 0.096/0.326 = 0.294

molefraction of CO2 = 0.084/0.326 = 0.26

molefraction of SF6 = 1-(0.294+0.26) = 0.446

partial pressure of CO = xCO*pTotal

     = 0.294*2356 mm of Hg

     = 693 mm of Hg

partial pressure of CO2 = xCO2*pTotal

          = 0.26*2356

          = 612.6 mm of Hg
           

partial pressure of SF6 = 2356 - (693+612.6)

   = 1054 mm of Hg