Question

A football bladder contains equimolar proportions of H2 and O2 .the composition by mass of the mixture effusing out of punctured football will be

Answer 1

Answer : The composition by mass of the mixture $H_2$ and $O_2$ effusing out of punctured football is in the ratio  4 : 1.

Solution :

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

$Rate\propto \frac{1}{\sqrt{\text{ Molar mass}}}$        ...........(1)

$Rate=\frac{W}{t}=\frac{Mass}{Time}$              ..............(2)

By equating equation (1) & (2), we get

$\frac{W_1}{W_2}=\sqrt{\frac{M_2}{M_1}}$         ..........(3)

where,

$W_1$ = Mass of $H_2$

$W_2$ = Mass of $O_2$

$M_1$ = Molar mass of  $H_2$

$M_2$ = Molar mass of  $O_2$

t = time

The molar mass of $H_2$ and $O_2$ are 2 & 32 g/mole respectively.

Now put all the values in above formula (3), we get

$\frac{W_1}{W_2}=\sqrt{\frac{32g/mole}{2g/mole}}=\sqrt{\frac{16}{1}}=\frac{4}{1}$

Hence, the composition by mass of the mixture $H_2$ and $O_2$ effusing out of punctured football is in the ratio 4 : 1.

Answer 2

Answer:

what is by equimolar

Explanation: