Question

X mL of H₂ gas effuses through a hole in a container in 5
seconds. The time taken for the effusion of the same volume
of the gas specified below under identical conditions is :
(a) 10 seconds : He (b) 20 seconds : O₂
(c) 25 seconds : CO (d) 55 seconds : CO₂

Answer 1

answer : option (b) 20 second : O2

X mL of H₂ gas effuses through a hole in a container in 5seconds.

Graham's law of diffusion, rate of effusion or diffusion is inversely proportional to square root of their density or molecular weight.

i.e., r ∝ 1/√d ∝ 1/√M

⇒r ∝ 1/√M

molecular weight of H2 = 2 g

molecular weight of O2 = 32g

here we see, ratio of molecules weight of H2 and O2 , M_{H2}/M_{O2} = 2/32 =1/16

√{M_{H2}/M_{O2} = 1/4

so, rate of effusion of H2/rate of effusion of O2 = 1/√{M_{H2}/M_{O2} =1/( 1/4) = 4

⇒(X mL/5sec)/(X mL/t sec) = 4

⇒t/5 = 4

⇒t = 20sec

hence, 20 second : O2

option (b) is correct choice.

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Answer 2

The time taken for the effusion of the same volume  of the gas specified below under identical conditions is 20 seconds : O₂

Step by step explanation:

Graham’s law of diffusion :

"It states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight."

The molecular formula is as follows.

                  $\bold{\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{1}}{M_{2}}}}$........................(1)

r = Rate of diffusion of gas

M = Molar mas of gas

Substitute the given values in equation (1)

               $\frac{r_{1}}{r_{2}}=\frac{t_{2}}{5}=\sqrt{\frac{M_{2}}{2}}$

                  $\Rightarrow t_{2}=20 sec$

                 $\Rightarrow M_{2}=32$

Therefore, the gas is Oxygen (molar mass = 32 g/mol).

Hence, correct option is b.

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