Q. A mixture of $CH_{4},N_{2}$ and $O_{2}$ is enclosed in a vessel of one litre capacity at $0{^\circ}C$. The ratio of particle pressures of gases is 1 : 4 : 2. Total pressure of the gaseous mixture is 2660 mm. the number of molecules of oxygen present in the vessel is
A) $\frac{6.02 \times 10^{23}}{22.4}$
B) $6.02 \times 10^{23}$
C) $22.4 \times 10^{22}$
D) 1000
Explanation
(a)
Particle pressure of oxygen $= \frac{2}{1 + 4 + 2} \times 2660$
$= 760\ mm$
Thus, 1 L oxygen gas is present at $0{^\circ}C$ and 760 mm pressure.
$\therefore$ Number of oxygen molecules $= \frac{6.023 \times 10^{23}}{22.4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.