Question

A mixture of argon and krypton gases, in a 5.38 L flask at 67 °C, contains 6.18 grams of argon and 9.66 grams of krypton. The partial pressure of krypton in the flask is________atm and the total pressure in the flask is______atm?

Answer 1

Answer:

The partial pressure of krypton in the flask is 0.59 atm and the total pressure in the flask is 1.39 atm

Explanation:

This must be solved with the Ideal Gas Law equation.

First of all we need the moles or Ar and Kr in the mixture

Moles = Mass / Molar mass

Molar mass Ar 39.95g/m

Moles Ar = 6.18 g/39.95 g/m → 0.154 moles

Molar mass Kr 83.8 g/m

Moles Kr = 9.66 g/ 83.8g/m → 0.115 moles

Total moles in the mixture: 0.154 moles + 0.115 moles = 0.269moles

Now, we have the total moles, we can calculate the total pressure.

P . V = n . R . T

(T° in K = T° in C + 273)

P. 5.38L = 0.269mol . 0.082 L.atm/mol.K . 340K

P = (0.269mol . 0.082 L.atm/mol.K . 340K) / 5.38 L

P = 1.39 atm

Now we have the total pressure, we can apply molar fraction so we can know the partial pressure of Kr.

Kr pressure / Total Pressure = Kr moles / Total moles

Kr pressure / 1.39 atm = 0.115 moles / 0.269 moles

Kr pressure = (0.115 moles / 0.269 moles) / 1.39atm

Kr pressure = 0.59 atm