Question

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?

Answer 1

Partial pressure of krypton= 167.12 mmHg

Partial pressure of argon = 565.88 mmHg

Step-by-step 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