Question

Calculate the volume occupied by 0.8 g of methane at STP.

Answer 1

Concept used:

According to the mole concept, one mole is the amount of the gas which has a volume of 22.4 L at STP.

Also, 1 mole is equal to Gram molecular weight

Therefore, Gram molecular weight= 22.4 L of the substance at STP

Step-1: Calculation of the gram molecular weight of Methane (CH₄)

Gram atomic weight of Carbon (C)= 12 g mol^-1

Gram atomic weight of Hydrogen (H)= 1 g mol^-1

Gram molecular weight of Methane (CH₄)= Gram atomic weight of Carbon (C)+ Gram atomic weight of Hydrogen (H)

Gram molecular weight of Methane (CH₄) = 12+ 4(1)

Gram molecular weight of Methane (CH₄) =16 g mol^-1

Step-2: Calculation of the volume occupied by 0.8 g of methane at STP.

16 g of Methane occupies volume= 22.4 L at STP

1 g of Methane occupies volume= $\frac{22.4}{16}$ L at STP

0.8 g of Methane occupies volume= $\frac{22.4}{16}\times 0.8=1.12$ L at STP

Therefore, the volume occupied by 0.8 g of methane at STP is 1.12 L.