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50 litre of dry $N_2$ is passed through 36g $H_2O$ at $27^{\large\circ}C$. After passage of gas , there is a loss of 1.20g in water . Calculate vapour pressure of water.

$(a)\;249.5mm\qquad(b)\;2.495mm\qquad(c)\;0.249mm\qquad(d)\;24.95mm$

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The water vapour occupies the volume of $N_2$ gas ,i.e 50 litre
$\therefore$ For $H_2O$ vapour
V = 50 litre
w = 1.20g
T = 300K
m = 18
PV = $\large\frac{w}{m} RT$
$P\times50 = \large\frac{1.2}{18}\times0.0821\times300$
P = 0.03284 atm
= 24.95 mm
Hence answer is (D)
answered Mar 26, 2014 by sharmaaparna1
 

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