Question

Calculate the molecular weight of a gas $X$ which diffuses four times as fast as another gas $Y$, which in turn diffuses twice as fast as another gas $Z$. Molecular weight of the gas $Z$ is 128. then X is :

Answer 1

$r_y=2r_z$
$\frac{r_y}{r_z}=\frac{2r_z}{r_z}=\sqrt{\frac{128}{M_y}}$
$4r_y=\frac{128}{4}=32$
$r_x=4r_y$
$\frac{r_x}{r_y}=\frac{4r_y}{r_y}=\sqrt{\frac{32}{M_x}}$
$M_x=2$