A $3.00 L$ oxygen gas is collected over water at $27^{o} C$ when the barometric pressure is $787 T o r r$ . If vapour pressure of water is $27 T o r r$ at $27^{o} C$ , moles of $O_{2}$ gas will be?

0.122

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0.126

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0.004

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0.152

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Solution

The correct option is A $0.122$
Given Volume (V) $= 3.00$ L
Temperature(T) $= 27^{0} C = 300 K$
Barometric pressure $= 787$ torr
Vapour pressure of water $= 27$ torr
total pressure $= 787 - 27 = 760 t o r r = 1 a t m$
Applying in ideal gas equation
PV=nRT
$n = \frac{P V}{R T}$
$n = \frac{1 \times 3}{0.0821 \times 300} = 0.122$

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