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3 years ago

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A sample of oxygen gas was collected via water displacement. since the oxygen was collected via water displacement, the sample i

s saturated with water vapor. if the total pressure of the mixture at 26.4 °c is 817 torr, what is the partial pressure of oxygen

1 answer:

NeX [460]3 years ago

3 0

Missing question: <span> The vapor pressure of water at 26.4 °C is 25.81 mm Hg.
p(total) = p(H</span>₂O) + p(O₂).
p(total) = 817 torr; <span> total pressure of the mixture.
</span>p(H₂O) = 25.81 torr, <span>partial pressure of water.
</span>p(O₂) = 817 torr - 25.81.
p(O₂) = 791.19 torr; partial pressure of oxygen.

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Explanation:

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Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

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Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

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Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

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The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

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