Question

Calculate the weight of the following gas which occupy a volume of 3.36 dm³ at STP.

helium

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= Gram atomic weight

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

Or Gram molecular weight= 22.4 dm³ of the substance at STP (1 litre= 1 dm³)

Step-1: Gram atomic weight of Helium (He)

Gram atomic weight of Helium (He)= 4 g mol^-1

(As Helium is a noble gas, it exists in the form of atoms only and not molecules)

Step-2: Calculation of weight in a volume of 3.36 dm³ at STP

For Helium gas, Atomic weight is 4 g.

22.4 dm³ of Helium gas (He) weighs= 4 g

1 dm³ of Helium gas (He) weighs= $\frac{4}{22.4}$ g

3.36 dm³ of Helium gas (He) weighs= $\frac{4}{22.4}\times 3.36=\frac{13.44}{22.4}=0.6$ g

Therefore, the weight of the Helium gas which occupy a volume of 3.36 dm³ at STP is 0.6 g.