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(C) According to Avogadro's Law,at constant temperature and pressure,the number of moles $(n)$ of gas is proportional to the volume $(V)$ of the conta

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Double that of methane

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(C) According to Avogadro's Law,at constant temperature and pressure,the number of moles $(n)$ of gas is proportional to the volume $(V)$ of the container. Since the volume of the flask is constant,the number of moles of $CH_4$ and $O_2$ will be equal.$n_{CH_4} = n_{O_2}$$\frac{w_{CH_4}}{M_{CH_4}} = \frac{w_{O_2}}{M_{O_2}}$Given $M_{CH_4} = 16 \ g/mol$ and $M_{O_2} = 32 \ g/mol$,we have:$\frac{w_{CH_4}}{16} = \frac{w_{O_2}}{32}$$w_{O_2} = 2 	imes w_{CH_4}$Thus,the mass of oxygen is double the mass of methane.

$A$ flask is weighed when filled with methane $(CH_4)$. The methane is then removed,and the flask is weighed again. When the flask is filled with oxygen $(O_2)$ at the same temperature and pressure,the mass of oxygen will be:

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