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nikklg [1K]
3 years ago
8

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
Chemistry
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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Answer : The correct option is, (c) 79.62

Explanation :

The formula used for percent humidity is:

\text{Percent humidity}=\text{Relative humidity}\times \frac{p-p^o_v}{p-p_v}\times 100   ..........(1)

The formula used for relative humidity is:

\text{Relative humidity}=\frac{p_v}{p^o_v}       ...........(2)

where,

p_v = partial pressure of water vapor

p^o_v = vapor pressure of water

p = total pressure

First we have to calculate the partial pressure of water vapor by using equation 2.

Given:

p=750mmHg

p^o_v=17.5mmHg

Relative humidity = 80 % = 0.80

Now put all the given values in equation 2, we get:

0.80=\frac{p_v}{17.5mmHg}

p_v=14mmHg

Now we have to calculate the percent humidity by using equation 1.

\text{Percent humidity}=0.80\times \frac{750-17.5}{750-14}\times 100

\text{Percent humidity}=79.62\%

Therefore, the percent humidity is 79.62 %

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3 years ago
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Explanation:

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