Dalton's law

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Question

A gaseous mixture contains 8 grams of oxygen, 14 grams of Nitrogen, 9 grams of water vapour. Partial pressure of water vapour in the mixture is (Assume water vapour behaves like ideal gas)

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Solution

$n_{N_{2}} = \frac{14}{28} = \frac{1}{2}; n_{O_{2}} = \frac{8}{32} = \frac{1}{4}; n_{H_{2} O} = \frac{9}{18} = \frac{1}{2}$
$X_{H_{2} O} = \frac{n_{H_{2} O}}{n_{N_{2}} + n_{O_{2}} + n_{H_{2} O}} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{1}{4} + \frac{1}{2}} = \frac{1}{2} \times \frac{4}{5} = \frac{4}{10}$

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Dalton's law

Remember concepts with our Masterclasses.

A gaseous mixture contains 8 grams of oxygen, 14 grams of Nitrogen, 9 grams of water vapour. Partial pressure of water vapour in the mixture is (Assume water vapour behaves like ideal gas)

nN2=1428=12;nO2=832=14;nH2O=918=12XH2O=nH2OnN2+nO2+nH2O=1212+14+12=12×45=410

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