Question

Verified by Toppr

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

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

$β΄r_{P}=0.567Γr_{Ne}$

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

$r=Mβ1β$

$β΄r_{Ne}=M_{Ne}β1β=20.18β1β=0.222$

Substituting the value of $r_{Ne}$ in $eq_{n}(1)$, we have

$r_{P}=0.222Γ0.567=0.126$.

From the formula,

$r_{P}=M_{P_{n}}β1β$

$M_{P_{n}}β=r_{P}1β=0.1261β=7.93$

$βM_{P_{x}}=(7.93)_{2}=62.88$

$n(P)=31g/mol62.88g/molββ2$

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

0

0