A) 1.774×10 -4 mol of anunidentified gaseous substance effuses through a tiny ho

Question

A) 1.774×10-4 mol of anunidentified gaseous substance effuses through a tiny hole in 97.7s.
Under identical conditions, 1.816×10-4 mol ofargon gas takes 82.9 s to effuse.
What is the molar mass of the unidentified substance (ing/mol)? B) The percentage C by mass in the above unidentifiedsubstance is 82.7%. It may also contain H and/or O, but no otherelement.
What is the molecular formula of the substance? C) Under identical conditions, how many moles of allene(C3H4) gas would effuse in 88.8 s?
A) 1.774×10-4 mol of anunidentified gaseous substance effuses through a tiny hole in 97.7s.
Under identical conditions, 1.816×10-4 mol ofargon gas takes 82.9 s to effuse.
What is the molar mass of the unidentified substance (ing/mol)? B) The percentage C by mass in the above unidentifiedsubstance is 82.7%. It may also contain H and/or O, but no otherelement.
What is the molecular formula of the substance? C) Under identical conditions, how many moles of allene(C3H4) gas would effuse in 88.8 s?

Explanation / Answer

Graham's law, also known as Graham'slaw of effusion, was formulated by Scottish physicalchemist ThomasGraham. Graham found experimentally that the rate of effusion of a gas isinversely proportional to the square root of the mass of itsparticles. This formula can be written as:

where:

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