Question

A football bladder contains equimolar proportions of $H_2$ and $O_2$. The composition by mass of the mixture effusion out of punctured football is in the ratio $(H_2:O_2)$:

• A. $1:4$
• B. $2\sqrt{2}:1$
• C. $1:2\sqrt{2}$
• D. $4:1$

Answer 1

The correct option is D $4:1$

The rate of effusion is inversely proportional to the square root of molar mass.
$r=\frac{w}{t}=\sqrt{\frac{1}{M}}$

$\frac{w}{w^{\prime }}=\sqrt{\frac{M^{\prime }}{M}}$

Here, $w$ and $w^{\prime }$ represents the mass of the hydrogen and oxygen respectively in the mixture effusion out of punctured football and $M$ and $M^{\prime }$ represents the molar mass of hydrogen and oxygen respectively.

$\frac{w}{w^{\prime }}=\sqrt{\frac{32}{2}}=4$

Hence, the ratio of the mass of hydrogen to the mass of oxygen in the mixture effusion out of punctured football is 4:1.