Question

1 L of gaseous mixture is effused in 5 min 11s, while 1 L of oxygen takes 10 min. The gaseous mixture contains methane and hydrogen .

Calculate: a. The density of gaseous mixture
b. the percentage by volume of each gas in the mixture

Answer 1

Given:

$d_1=16$

$t_1=10\times 60=600s$

$d_2=?$

$t_2=5\times 60+11=311s$

Volume is same, so

$\frac{t_2}{t_1}=\sqrt{\frac{d_2}{d_1}}$

$\frac{{311}^2}{{600}^2}=\sqrt{\frac{d_2}{16}}$

$d_2=4.3$

$d_{H_2}=1$

and $d_{C{H}_4}=8$

So

$d_{mix}=\frac{d_{C{H}_4}\times \left(x\right)+(100-x)d_{H_2}}{100}$

$4.3\times 100=8x+(100-x)\times 1$

$430=8x+100-x$

$330=7x$

$x=47.14$%

%$C{H}_4=47.14$, %$H_2=52.86$