The mass of oxygen gas occupying a volume of | vedclass

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(B) Using the Ideal Gas Law: $PV = nRT = \frac{m}{M}RT$Given: $P = 760 	ext{ mm of Hg} = 1 	ext{ atm}$,$V = 11.2 	ext{ L}$,$T = 27^{\circ}C = 300 \

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$0.01456$

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(B) Using the Ideal Gas Law: $PV = nRT = \frac{m}{M}RT$Given: $P = 760 	ext{ mm of Hg} = 1 	ext{ atm}$,$V = 11.2 	ext{ L}$,$T = 27^{\circ}C = 300 	ext{ K}$,$M = 32 	ext{ g/mol}$,$R = 0.0821 	ext{ L atm K}^{-1} 	ext{ mol}^{-1}$.Rearranging for mass $m$: $m = \frac{PVM}{RT}$$m = \frac{1 	imes 11.2 	imes 32}{0.0821 	imes 300}$$m = \frac{358.4}{24.63} \approx 14.55 	ext{ g}$Converting to kilograms: $m = \frac{14.55}{1000} \approx 0.01456 	ext{ kg}$.

The mass of oxygen gas occupying a volume of $11.2 \text{ litres}$ at a temperature $27^{\circ}C$ and a pressure of $760 \text{ mm}$ of mercury in kilograms is $[$Molecular weight of oxygen $= 32]$

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