JEE Main · 2010 · 11 Apr 2010 · Offline · States of Matter

1 0^{- 4} dm^{3}

of water is introduced into a

1.0 dm^{3}

flask to 300 K , how many moles of water are in the vapour phase when equilibrium is established ?
(Given : Vapour pressure of

H_{2} O

at 300 K is

3170 Pa; R = 8.314 J K^{- 1} mol^{- 1}

)

Answer: (D)

Solution

At equilibrium the vapour is saturated, so its pressure equals the vapour pressure of water at

300

3170

Pa, as long as some liquid remains. Treating the vapour as ideal, the moles present are

n = \frac{P V}{R T} = \frac{3170 \times 1.0 \times 1 0 ^{- 3}}{8.314 \times 300} \approx 1.27 \times 1 0^{- 3}

mol. The

1 0^{- 4}

^{3}

of liquid introduced is about

5.6 \times 1 0^{- 3}

mol, more than enough to supply this vapour, leaving

1.27 \times 1 0^{- 3}

mol in the vapour phase.

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option