A 3.0-l sample of helium was placed in a container fitted with a porous membrane. half of the helium effused through the membrane in 24 hours. a 3.0-l sample of oxygen was placed in an identical container. how many hours will it take for half of the oxygen to effuse through the membrane?

Answer: M in this equation is molar mass. If A is He and B is O2 then MA = 4 g/mol and MB = 32 g/mol rate A = 1.5L/24hr rate B = 1.5L/?hr and rateA/rateB = ?/24

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ardni313 ardni313 · Answer 2 · 2020-01-19T11:18:02+01:00

It will take 68.182 hours for half of the oxygen to effuse through the membrane

Further explanation

Graham's law: the rate of effusion of a gas is inversely proportional to the square root of its molar masses or

the effusion rates of two gases = the square root of the inverse of their molar masses:

[tex]\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }[/tex]

or

[tex]\rm M_1\times r_1^2=M_2\times r_2^2[/tex]

[tex]\rm r_{He}=\dfrac{1.5\:L}{24\:Hr}=\dfrac{1}{16}\:L/Hr\\\\r_{He}^2\times M_{He}=r_O_2^2\times M_{O_2}[/tex]

[tex]\rm M_{He}=4,M_{O_2}=32\\\\(\dfrac{1}{16})^2\times 4=r_{O_2}^2\times 32\\\\r_{O_2}=\sqrt{\frac{1}{2048} }\\\\r_{O_2}=0.022\:L/Hr[/tex]

For 1.5 L O₂ it will take :

1.5 L : 0.022 L/Hr = 68.182 hours

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