Q. If $10^{- 4}\ {dm}^{3}$ of water is introduced into a $1\ {dm}^{3}$ flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established (Given vapour pressure of $H_{2}O$ at 300K is 3170 $Pa;R = 8.314\ {JK}^{- 1}\ {mol}^{- 1}$)
A) $5.56 \times 10^{- 6}\ mol$
B) $1.53 \times 10^{- 2}\ mol$
C) $4.46 \times 10^{- 2}\ mol$
D) $1.27 \times 10^{- 3}\ mol$
Explanation
(d)
$n = \frac{pV}{RT} = \frac{3170 \times 10^{- 3}}{8.314 \times 300} = 1.27 \times 10^{- 3}\ mol$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.