mix 45 grams of oxygen gas with 183 grams of argon gas in a volume of 1697mL at 160

We mix 45 grams of oxygen gas with
183 grams of argon gas in a volume of 1697mL
at 160 C. What will be the final pressure of
the gas mixture?
Answer in units of atm




Okay, first of all, let's convert all units to SI units.

1.000mL= 1.000x10⁻⁶m³ 
1697mL= (1.000x10⁻⁶)(1697)
= 1.697x10⁻³m³ 
160°C= (160+273)K
= 433K

Relative Molecular Mass(RMM) of O₂
= 2(RAM of O)
= 2(16)
= 32

32g= 1 mole of O₂
45g= 45/32
= 1.40625 moles of O₂

Relative Atomic Mass(RAM) of Ar
= 39.95

39.95g= 1 mole of Ar
183g= 183/39.95
= 4.580725907 moles of Ar 

Total number of moles of O₂ and Ar
= 5.986975907

We shall now use the ideal gas equation PV= nRT where P is the total pressure exerted by the gas particles, V is the volume of the container, n is the total number of moles of gas particles, R is the molar gas constant(8.31J/mol/K) and T is the temperature.

P= nRT/V
= (5.986975907)(8.31)(433)/1.697x10⁻³
= 1.269447043x10⁷N/m²

1.01x10⁵N/m²= 1.00atm
1.269447043x10⁷N/m²= 1.269447043x10⁷/1.01x10⁵
= 125.687826atm
= 126atm correct to 3sf

Thus, the final pressure will be 126atm. I hope this helps and feel free to send me an e-mail if you have any doubts!