Question

One mole of nitrogen gas at $0.8atm$ takes 38 seconds to diffuse through a pin hole whereas one mole of an unknown compound of oxygen with fluorine at $1.6$ atm takes 56.26 seconds to diffuse through the same hole, then the molecular formula of the compound is:

(Atomic mass of xenon: 131.3 u)

• A. $Xe{F}_2$
• B. $Xe{F}_4$
• C. $Xe{F}_5$
• D. $Xe{F}_6$

Answer 1

Answer: (A)

$Xe{F}_2$

Given : $n_1=1mol,P_1=0.8atm,t_1=38s$ & $n_2=1mol,P_2=1.6atm,t_1=56.26s$

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\times \frac{P_1}{P_2}\frac{n_1}{t_1}\times \frac{t_2}{n_2}=\sqrt{\frac{M_2}{M_1}}\times \frac{P_1}{P_2}$

Or substitution,

$\frac{1}{38}\times \frac{56.26}{1}=\sqrt{\frac{M_2}{28}}\times \frac{0.8}{1.6}{M}_2=252g$

Let the compound be $Xe{F}_x$

$131.3+19x=252x=\frac{252-131.3}{19}\simeq 6$

$\therefore$ The compound is $Xe{F}_6.$