$O_{2}$ is collected over water a t 20C.The pressure inside shown by the gas is 750 mm Hg. What is the pressure due to $O_{2}$ it vapour pressure of $H_{2}$ O is 18 mm at 20'C?

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Solution

Here the total pressure of the gas in the container is $750 m m H g$
Here vapour pressure of the $O_{2}$ is $18 m m H g$
Thus, according to Dalton's law,
$P_{T} = P_{O_{2}} + P_{H_{2} O}$
$∴ P_{O_{2}} = P_{T} - P_{H_{2} O} = 750 - 18 = 732 m m H g$

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O2 is collected over water a t 20C.The pressure inside shown by the gas is 750 mm Hg. What is the pressure due to O2 it vapour pressure of H2O is 18 mm at 20'C?

Here the total pressure of the gas in the container is 750 mm HgHere vapour pressure of the O2 is 18 mm HgThus, according to Dalton's law,PT=PO2+PH2O∴PO2=PT−PH2O=750−18=732 mm Hg

$O_{2}$ is collected over water a t 20C.The pressure inside shown by the gas is 750 mm Hg. What is the pressure due to $O_{2}$ it vapour pressure of $H_{2}$ O is 18 mm at 20'C?

Here the total pressure of the gas in the container is $750 m m H g$
Here vapour pressure of the $O_{2}$ is $18 m m H g$
Thus, according to Dalton's law,
$P_{T} = P_{O_{2}} + P_{H_{2} O}$
$∴ P_{O_{2}} = P_{T} - P_{H_{2} O} = 750 - 18 = 732 m m H g$