Question

H2 and O2 as kept in the mass ratio 1:8 respectively at 6 atm. if small orifice is made the relative rate of effusion of H2 with respect to O2 initially is?

Answer 1

Explanation:

rate of diffusion =

r

2

r

1

=

V

2

V

1

×

M

1

M

2

=

1

2

×

2

32

=

1

8

Answer 2

Given:

The mass ratio of H₂ AND O₂ = 1:8

Pressure = 6atm

To find:

The relative rate of effusion of H2 with respect to O2 initially

Calculation:

We know the formula,

  $\frac{r_{1} }{r_{2} } = \frac{n_{1} }{n_{2} }\sqrt{\frac{M_{2} }{M_{1} }}$

By substituting the given values in the formula, we get

  $\Rightarrow \frac{r_{1} }{r_{2} } = \frac{n_{1} }{n_{2} }\sqrt{\frac{M_{2} }{M_{1} }} \\\\\Rightarrow \frac{r_{1} }{r_{2} } = \frac{1 }{8 }\sqrt{\frac{32 }{2 }}\\\\\Rightarrow \frac{r_{1} }{r_{2} } = \frac{1 }{8}\times4\\\\\Rightarrow \frac{r_{1} }{r_{2} } = \frac{1 }{2}$

The relative rate of effusion of H2 with respect to O2 initially is 1:2