a) 56g
Calculation:
At STP,
22.4 L of N₂ = 1 mol
We have given 44.8 L of N₂, therefore,
44.8 L of N₂ = [tex]\frac{44.8}{22.4}[/tex]
= [tex]2[/tex] mol
We know that,
1 mol of N₂ = 28 g
Hence,
2 mol of N₂ = 28 × 2
= 56g
Hence, there are 56 g of N₂ in 44.8 L of nitrogen gas.
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