Question

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

Answer 1

• The given volume of nitrogen gas(N₂) =3.36 dm³ = 3.36 L
• We know that 1 mole of nitrogen gas(N₂) = 6.022 × 10²³ molecules = 22.4L volume at STP
• Use the unitary method to calculate the number of moles of nitrogen gas for the given volume,

22.4 L volume of nitrogen gas (N₂) has a number of moles= 1 mol

Then, 1 Litre volume of nitrogen gas (N₂) has a number of moles = $\frac{1}{22.4}$

Therefore 3.36 L of nitrogen gas (N₂) has a number of moles = $\frac{1}{22.4}$× 3.36 = 0.15 moles.

• 1 mole of Nitrogen gas (N₂) has weight = molecular weight of nitrogen gas (N₂) = 28 g

There fore 0.15 mole nitrogen gas (N₂) has weight = 0.15 × 28 = 4.2g

Thus the weight of nitrogen gas(N₂) which occupy a volume of 3.36 dm³ at STP is 4.2g.