Question

A mixture containing 2 moles of He and 1 mole of CH4

is taken in a closed container and made to

effuse through a small orifice of container. Then, which is the correct effused volume percentage of

He and CH4

initially, respectively :

(A) 40%, 60% (B) 20% , 80% (C) 80% , 20% (D) 60% , 40%​

Answer 1

Answer:

Molar mass of CH4=16g/mol

Molar mass of He=4

Explanation:

Graham law states

Rate of diffusion of substanceinversely relate to it molar mass

R1=He

R2=CH4

R1/R2=(M2) ^1/2/(M1)1^2

R1/R2=(16^1/2)/(4^1/2)

R1/R2=4/2

R1/R2=2/1

(C) is the correct option

Answer 2

Answer:

The rate of effused volume percentage of $He$ is $80\%$.

The rate of effused volume percentage of $CH_{4}$ is $20\%$.

Therefore, option c) $80\%, 20\%$ is correct.

Explanation:

Given,

The number of moles of $He$, $n_{1}$ = $2 moles$

The number of moles of $CH_{4}$, $n_{2}$ = $1mole$

The effused volume percentage rate of $He$?

The effused volume percentage rate of $CH_{4}$?

Now,

• Let the effusion rate of $He$ = $R_{1}$
• Let the effusion rate of $CH_{4}$ = $R_{2}$

Also,

• The molar mass of $He$, $M_{1}$ = $4g/mol$
• The molar mass of $CH_{4}$, $M_{2}$ = $16g/mol$

As we know,

• The rate of effusion of the gases is directly proportional to the moles of the gases and inversely proportional to the square root of their molar mass.

Therefore,

• $\frac{R_{1} }{R_{2} }= \frac{n_{1} }{n_{2} } \sqrt{\frac{M_{2} }{M_{1} } }$

After putting all the values in the equation, we get:

• $\frac{R_{1} }{R_{2} }= \frac{2 }{1 } \sqrt{\frac{16 }{4} }$
• $\frac{R_{1} }{R_{2} }= \frac{4}{1}$

The ratio of the effusion rate of $He:CH_{4}$ is $4:1$.

Now, the total effused volume rate is $5$.

Hence, the rate of effused volume percentage of $He$ = $\frac{4}{5}\times 100$ = $80\%$.

And the rate of effused volume percentage of $CH_{4}$ = $\frac{1}{5}\times 100$ = $20\%$.