Answer to Question #147990 in General Chemistry for Laura

Question

What is the pressure in
mm of Hg , of a gas mixture
that contains 1g of H2, and 8.0
g of Ar in a 3.0 L container at
27°C.

Answer 1

Mass of H₂ = 1g

Molar mass of H₂ = 2gmol^-1

moles of H₂ = 1/2 = 0.5mol

Mass of Ar = 8g

Molar mass of Ar = 40gmol^-1

moles of Ar = 8/40 = 0.20mol

The partial pressure exerted by Hydrogen P_H is

P_H = nRT/V

Where n= 0.5mol, R= 0.0821L.atm/K.mol, T= 27+273 = 300K, V= 3.0L

P_H = 0.5 x 0.0821 x 300/3.0= 4.105atm

The partial pressure exerted by Argon is

P_Ar= nRT/V

Where n= 0.2mol, R= 0.0821L.atm/K.mol, T=300K, V= 3.0L

P_Ar= 0.2x0.0821x300/3.0 = 1.642atm

Total pressure= P_H + PAr

P_T= 4.105 + 1.642

P_T= 5.747atm

1atm= 760mmHg

5.747atm= 5.747x760= 4367.72mmHg