Answer: 67.23 L
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Putting in the values we get:
[tex]\text{Number of moles of methane}=\frac{48g}{16g/mol}=3moles[/tex]
1 mole of methane occupies = 22.4 L
Thus 3 moles of methane occupy = [tex]\frac{22.4}{1}\times 3=67.23L[/tex]
Thus volume of 48 grams of methane gas is 67.23 L
