Question

A mixture if Krypton and Neon gases, at a total pressure of 678mm Hg, contains 5.96 grams of krypton and 4.42 grams of neon what is the partial pressure of each gas in the mixture in mm Hg?

Answer 1

First we need to calculate the mole fraction of each gas in the mixture, with molar mass we can determine that:

5.96 g Kr * (1mol/ 83.798 g) = 0.0711 mol Kr

4.42 g Ne *(1mol/ 20.1797 g) = 0.219 mol Ne

Now the mole fraction could be calculated using the following formula:

For Krypton, we have:

And for the neon we have 0,755 for the mole fraction.

The partial pressure of an individual gas component in an ideal gas can be obtained using this expression:

where: x_i is the mole fraction of any individual gas component in a gas mixture, pi is the partial pressure of any individual gas component in a mixture and p is the total pression of the gas mixture.

The partial pressure for Krypton will be:

p_Kr = 0.254*678 mmHg = 166 mmHg

And the partial pressure for Neon will be:

pKr = 0,755*678 mmHg = 512 mmHg