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Osmotic Pressure

At STP, a con...

Question

At STP, a container has $1$ mole of $A r$ , $2$ moles of $C O_{2}, 3$ moles of $O_{2}$ and $4$ moles of $N_{2}$ . Without changing the total pressure of $O_{2}$ is removed, the partial pressure of $O_{2}$ is:

changed by about

16

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halved.

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changed by

26

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unchanged.

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Solution

The correct option is B changed by $26$ %.
Mole fraction of $A = \frac{moles of A}{total moles}$
Given,
Moles of $A r = 1$ , moles of $C O_{2} = 2$ ,
moles of $O_{2} = 3$ , moles of $N_{2} = 4$ ,
moles of $O_{2}$ removed $= 1$

Mole fraction of $O_{2}$ at initial stage

$= \frac{3}{1 + 2 + 3 + 4} \times 100 = \frac{3}{10} \times 100 = 30$

Mole fraction of $O_{2}$ at final stage

$= (\frac{3}{10} - \frac{2}{1 + 2 + 3 + 4}) \times 100$

$= (\frac{3}{10} - \frac{2}{9}) \times 100 = \frac{70}{9} \approx 8$

$∴$ % change $= \frac{8}{30} \times 100 = 26$ %

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At STP, a container has 1 mole of Ar, 2 moles of CO2,3 moles of O2 and 4 moles of N2. Without changing the total pressure of O2 is removed, the partial pressure of O2 is:

At STP, a container has $1$ mole of $A r$ , $2$ moles of $C O_{2}, 3$ moles of $O_{2}$ and $4$ moles of $N_{2}$ . Without changing the total pressure of $O_{2}$ is removed, the partial pressure of $O_{2}$ is: