When 10.33 g of neon is combined in a 23.76 L container at 94oC with 8.39 g of argon, what is the partial pressure (in mmHg) of argon?

Molar mass of Ar = 39.95 g/mol

Consider the Ar gas in the mixture :
No. of moles, n = 10.33/39.95 mol
Volume, V = 23.76 L
Temperature, T = (273 + 94) K = 367 K
Partial pressure = y mmHg = y/760 atm



Gas law : PV = nRT
Then, P = nRT/V

Partial pressure of Ar, P
= (10.33/39.95) × 0.0821 × 367 / 23.76 atm
= [760 × (10.33/39.95) × 0.0821 × 367 / 23.76] × 760 mmHg
= 249 mmHg