A mixture of 10.0 g of Ne and 10.0 g of Ar was kept in a 1.00 L container at 765 torr.Calculate the partial pressure of Ar inside the container.

Question

Quizplus · Accepted Answer

To find the partial pressure of Ar, we need to use the mole fraction of Ar in the mixture.

Moles of Ne = Mass of Ne / Molar mass of Ne = 10.0 g / 20.18 g/mol = 0.495 mol
Moles of Ar = Mass of Ar / Molar mass of Ar = 10.0 g / 39.95 g/mol = 0.250 mol

Total moles of gas in the container = 0.495 mol + 0.250 mol = 0.745 mol

Mole fraction of Ar = Moles of Ar / Total moles = 0.250 mol / 0.745 mol = 0.335

Partial pressure of Ar = Mole fraction of Ar x Total pressure 
= 0.335 x 765 torr 
= 257 torr

Therefore, the answer is B.

A Mixture of 10