Molar ratio in which He and CH4 should be mixed so that when mixture is allowed to effuse through pinhole, initially both gases comes out at equal rate?

Question

Molar ratio in which He and CH4 should be mixed so that when mixture is allowed to effuse through pinhole, initially both gases comes out at equal rate? (A) 4:1 (B) 1:4 (C) 1:2 (D) 2:1

Tardigrade Admin · Accepted Answer

rα (n/√m)⇒ (r(He)/r((CH)4))=(n(He)/n((CH)4)).√(16/4)=1

Q. Molar ratio in which $He$ and $C H_{4}$ should be mixed so that when mixture is allowed to effuse through pinhole, initially both gases comes out at equal rate?

$4 : 1$

$1 : 4$

$1 : 2$

$2 : 1$

$r α \frac{n}{m} \Rightarrow \frac{r _{(He)}}{r _{((C H)_{4})}} = \frac{n _{(He)}}{n _{((C H)_{4})}} . \frac{16}{4} = 1$

Q. Molar ratio in which He and CH4​ should be mixed so that when mixture is allowed to effuse through pinhole, initially both gases comes out at equal rate?

4:1

1:4

1:2

2:1