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Grade: 12th pass

                        

a mixture of H2 and O2 in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas comming out initially?

3 years ago

Answers : (1)

Ishika dammani
44 Points
							Answer.... 8 :1 , which means 8 molecules of H2 gas will effuse in every molecule of O2 gas. Graham`s Law: Rate1/Rate2 = p1/p2 √M2/M1 since p is constant therefore p1/p2 = 1 therefore, Rate1/Rate2 = √M2/M1 where, Rate 1, Rate2 are the rate of effusion of the gases involved. M1, M2 are the molar masses of the gases given: H2 = gas 1 O2 = gas 2 molar masses: H2 = 1g/mol * 2 = 2g/mol = M1 O2 = 16g/mol * 2 = 32g/mol = M2 rate given is 2:1 therefore, Rate H2/RateO2 = √32g/√2 = 4/1 rate = 2:1, therefore, 4*2 = 8 1*1 = 1 therefore, 8:1 , 8 H2 molecules will effuse for every molecule of O2. 
						
3 years ago
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