Question

A mixture of hydrogen and argon gases is maintained in a 8.48L flask at a pressure of 2.57 atm and a temperature of 64 C. If the gas mixture contains 0.481 grans of hydrogen, the number of grams of argon in the mixture is ____g.

Answer 1

Mass of hydrogen = 0.481 g

Molar mass of hydrogen = 2 g/mol

Moles of hydrogen (n_H2) = mass of hydrogen/molar mass of hydrogen

= 0.481 g/2 g/mol

= 0.240 mol

Let us say that the moles of Ar in the mixture = n_Ar

Total moles of the gas in the mixture (n)= (0.240 mol + n_Ar)

Volume of the gas (V) = 8.48 L

Pressure of the gas (P) = 2.57 atm

Temperature (T) = 64 ^oC = (64 + 273) K = 337 K

Universal gas constant = 0.08206 L-atm/mol-K

From ideal gas law,

PV = nRT

or, 2.57 atm x 8.48 L = (0.240 mol + n_Ar) x 0.08206 L-atm/mol-K x 337 K

or, (0.240 mol + n_Ar) = 0.788 mol

or, n_Ar = 0.788 mol - 0.240 mol

or, n_Ar = 0.548 mol

Thus, the moles of Ar in the mixture = 0.548 mol

Molar mass of Ar = 39.9 g/mol

Hence, the mass of argon in the mixture = 0.548 mol x 39.9 g/mol = 21.9 g