Question

Solid phosphorus melts and evaporates at high temperature Gaseous phosphorus effuses at a rate that is $0.567$ times that of Ne in the same apparatus under the same conditions . How many atoms are in a molecule of gaseous phosphorus.

Answer 1

Let $r_P$ and $r_{Ne}$ be th rate of effusion of phosphorous and neon respectively.

Given that phosphorus effuses at a rate that is $0.567$ times that of Ne in the same apparatus under the same conditions.

$\therefore r_P=0.567\times r_{Ne}$

As we know that, rate of effusion, 'r', of any gas of molar mass 'M' is given by-

$r=\frac{1}{\sqrt{M}}$

$\therefore r_{Ne}=\frac{1}{\sqrt{M_{Ne}}}=\frac{1}{\sqrt{20.18}}=0.222$

Substituting the value of $r_{Ne}$ in ${eq}^n\left(1\right)$, we have

$r_P=0.222\times 0.567=0.126$.

From the formula,

$r_P=\frac{1}{\sqrt{M_{P_n}}}$

$\sqrt{M_{P_n}}=\frac{1}{r_P}=\frac{1}{0.126}=7.93$

$\Rightarrow M_{P_x}={\left(7.93\right)}^2=62.88$

$n\left(P\right)=\frac{62.88g/mol}{31g/mol}\approx 2$

Hence there are $2$ atoms in a molecule of gaseous phosphorus.