Question

  1.5 liters of a gas mixture at 15°C and 900 kPa contains 421mg/ L of H₂gas (not H₂S) partial pressure exerted by this gas

Answer 1

486 kPa

Explanation

No of moles = mass/molar mass

No of moles of H2 in 1L = 0.421g/ 2.016g/mol = 0.202883

No of moles of H2 in 1.5L = 0.202883mol × 1.5 = 0.3043 mol

Ideal gas equation is

PV =nRT

P = 900kPa = 8.882atm

V = 1.5L

T = 15°C = 288.15K

R = 0.082057(L atm /mol K)

total no of moles, n = PV /RT

n = 8.882atm × 1.5L /(0.082057(L atm/mol K) × 288.15K)

= 0.5635 mol

Mole fraction of H2 = 0.3043mol/0.5635mol = 0.5400

partial pressure = mole fraction × Total pressure

partial pressure of H2 = 0.5400×900kPa = 486 kPa