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$O_2$ is collected over water at $20^{\large\circ}C$. The pressure inside shown by the gas is 740 mm of Hg. What is the pressure due to $O_2$ alone if V.P of $H_2O$ is 18 mm at $20^{\large\circ}C$

$(a)\;722mm\qquad(b)\;240mm\qquad(c)\;720mm\qquad(d)\;480mm$

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According to Dalton's law of partial pressure
Given
$P_M$ = 740 mm
$P_{H_2O}$ = 18 mm
$P_M or P_{wet\;O_2} = P_{O_2} + P_{H_2O}$
$740 = P_{O_2} + 18$
$P_{O_2} = 740 - 18$
$\;\;\;\;\;\;= 722 mm$
Hence answer is (A)
answered Mar 26, 2014 by sharmaaparna1
 

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