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If $10^{- 4} d m^{3}$ of water is introduced into a $1.0 d m^{3}$ flask to 300k, 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; $R = 8.314 J K^{- 1} m o l^{- 1}$ )

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_{$5.56 \times 10^{- 3} m o l$}

_{$5.53 \times 10^{- 2} m o l$}

_{$4.46 \times 10^{- 2} m o l$}

_{$1.27 \times 10^{- 3} m o l$}

answer is D.

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Detailed Solution

$n = \frac{P V}{R T}$

$= \frac{3170 \times 10^{- 5} a t m \times 1 L}{0.0821 L a t m k^{- 1} m o l^{- 1} \times 300 K}$

$= 1.27 \times 10^{- 3} m o l$

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