4 g of H 2 effused through a pinhole in 10 sec at constant temperature and pressure. The amount of oxygen effused in the same time interval and at the same conditions of temperature and pressure wou

Question

4 g of H 2 ​ effused through a pinhole in 10 sec at constant temperature and pressure. The amount of oxygen effused in the same time interval and at the same conditions of temperature and pressure would be :-

TestHub · Accepted Answer

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Rate $\propto \frac{1}{M}$

For two gases, H₂ and O₂, at the same temperature and pressure:

$\frac{R a t e _{H_{2}}}{R a t e _{O_{2}}} = \frac{M _{O_{2}}}{M _{H_{2}}}$

Given:

Mass of H₂ = 4 g

Time = 10 s

Molar mass of H₂ ( $M_{H_{2}}$ ) = 2 g/mol

Molar mass of O₂ ( $M_{O_{2}}$ ) = 32 g/mol

Rate of effusion is proportional to the amount (mass) effused in a given time.

$\frac{n _{H_{2}} / T im e}{n _{O_{2}} / T im e} = \frac{32}{2}$

$\frac{2}{n _{O_{2}}} = 16$

$n_{O_{2}} = \frac{2}{4} = \frac{1}{2}$

Mass of O₂ = 0.5 x 32 = 16g

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