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1.47 litre of a gas is collected over water at $30^{\large\circ}C$ and 744mm of Hg . If the gas weight 1.98g and vapour pressure of water at $30^{\large\circ}C$ is 32mm , What is the molecular weight of gas?

$(a)\;28.6g\qquad(b)\;30.76g\qquad(c)\;35.76\qquad(d)\;32.10g$

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Given
w = 1.98g
V = 1.47 litre
T = 303 K
P = 744 - 32 = 712mm = $\large\frac{712}{760}$ atm
Since For dry gas , $PV = \large\frac{w}{m}RT$
$=\large\frac{712}{760}\times1.47=\large\frac{1.98}{m}\times0.0821\times303$
$\therefore m = 35.76g$
Hence answer is (c)
answered Feb 20, 2014 by sharmaaparna1
 

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