Question

Solid phosphorus melts and vapourizes at a high temperature. Gaseous phosphorus effuses at a rate that is 0.57 times that of neon (Ne) in the same apparatus under the same conditions. How many atoms are in a molecule of gaseous phosphorus?
(Ne = 20; P = 31)

Answer 1

$r_1=$ rate of effusion of phosphorus
$r_2=$ rate of Ne
$\therefore \frac{r_1}{r_2}=0.567=\sqrt{\frac{M_2}{M_1}}$
or, 0.57 = $\sqrt{\frac{20}{x\times 31}}$ where $x$ is the number of phosphorus atoms.
squaring both sides
$\approx 0.325=\frac{20}{31x}$ or, $x=\frac{20}{31\times 0.325}\approx 2$
Hence, number of atoms present in phosphorus molecule is 2.