Question

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

Answer 1

• The given volume of Helium gas(He) =3.36 dm³ = 3.36 L
• We know that 1 mole of Helium gas(He) = 6.022 × 10²³ molecules = 22.4L volume at STP
• Use the unitary method to calculate the number of moles of the given Helium gas(He) for the volume,

22.4 L volume of Helium gas(He) has a number of moles= 1 mol

Then, 1 Litre volume of Helium gas(He)has number of moles = $\frac{1}{22.4}$

There fore 3.36 L of Helium gas(He) has a number of moles = $\frac{1}{22.4}$× 3.36 = 0.15 mole.

• 1 mole of Helium gas(He) has weight = Molecular weight of Helium gas(He) = 4 g

There fore 0.15 mole Helium gas(He) has weight = 0.15 × 4 = 0.6g

Thus the weight of Helium gas(He) which occupy a volume of 3.36 dm³ at STP is 0.6g.