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Bad White [126]
3 years ago
15

Estimate the molar mass of a gas that effuses at 1.80 times the effusion rate of carbon dioxide. answer in units of g/mol.

Chemistry
1 answer:
Effectus [21]3 years ago
3 0
From Grahams Law the rates of effusion of two gases are inversely proportional to the square roots of their molar masses at the same temperature and pressure.
Therefore; R1/R2 = √mm2/√mm1
The molecular mass of Carbon dioxide is 44 g
Hence;  1.8 = √(44/x
             3.24 = 44/x
                x = 44/3.24
                   = 13.58 
Therefore, the molar mass of the other gas is 13.58 g/mol
           
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518 mL

Explanation:

We can solve this using Boyle's Law Formula

P1V1 = P2V2

where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume

here , the initial pressure is 1 atm and the initial volume is 725mL

we are given the final pressure 1.4 and we need to find the final volume

so we have p1v1 = p2v2

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(1)(725) = (1.4)v2

==> multiply 1 and 725

725 = (1.4)(v2)

==> divide both sides by 1.4

v2 = 518

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