A mixture of krypton and argon gases, at a total pressure of 733 mm Hg, contains 3.97 grams of krypton and 6.34 grams of argon. What is the partial pressure of each gas in the mixture?

Question

Avah Holloway

Chemistry

29 November, 00:03

A mixture of krypton and argon gases, at a total pressure of 733 mm Hg, contains 3.97 grams of krypton and 6.34 grams of argon. What is the partial pressure of each gas in the mixture?

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Ayana · Answer 1

Partial pressure of krypton = 167.12 mmHg

Partial pressure of argon = 565.88 mmHg

Explanation:

The partial pressure of a gas in a mixture of gasses is equal to the total pressure multiplied (Pt) by the mole fraction of the gas (X):

P = X Pt

The total pressure is Pt = 733 mmHg

The mole fraction is given by the following:

X = number of moles of gas / total number of moles

For krypton (Kr), the molecular weight is 83.8 g/mol and we calculate the number of moles by dividing the mass into the molecular weight as follows:

moles of Kr = 3.97 g / (83.8 g/mol) = 0.047 moles

For argon (Ar), the molesular weight is 39.9 g/mol, so we calculate the number of moles as follows:

moles of Ar = 6.34 g / (39.9 g/mol) = 0.159

Now, we calculate the total number of moles (nt):

nt = moles of Kr + moles of Ar = 0.047 moles + 0.159 moles = 0.206 moles

The mole fraction of each gas is now calculated:

X (Kr) = moles of Kr/nt = 0.047 moles/0.206 = 0.228

X (Ar) = moles of Ar/nt = 0.159 moles/0.206 = 0.772

Finally, with the mole fractions and the total pressure we calculate the partial pressure of each gas as follows:

P (Kr) = X (Kr) x Pt = 0.228 x 733 mmHg = 167.12 mmHg

P (Ar) = X (Ar) x Pt = 0.772 x 733 mmHg = 565.88 mmHg