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If $10^{-4} dm^3$ of water is introduced into a $1.0dm^3$ flask to 300K. How many moles of water are in the vapour phase when equilibrium is established? (Given Vapour pressure of $H_2O$ at 300K is 3170 Pa ; $R = 8.314JK^{-1}mol^{-1}$

$(a)\;5.56\times10^{-3}mole\qquad(b)\;1.53\times10^{-2}mole\qquad(c)\;4.46\times10^{-2}mole\qquad(d)\;1.27\times10^{-3}mole$
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Given
Vapour pressure of $H_2O$ at 300K is 3170Pa
$R = 8.314JK^{-1}mol^{-1}$
Since PV = nRT
$\therefore n = \large\frac{3170\times10^{-5}atm\times1 L}{0.082LatmK^{-1}mol^{-1}\times300K}$
= $1.27\times10^{-3}mole$
Hence answer is (d)
answered Feb 14, 2014 by sharmaaparna1
 

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