Answer:
V(NH₃(g)) at STP = 0.8928 Liters ≅ 0.9 Liter (1 sig. fig)
Explanation:
Given 340g NH₃(g) => ?Volume at STP
Convert grams to moles by dividing by formula wt of ammonia
= 340g/17g/mole = 20 moles
1 mole of any gas at STP conditions (0°C, 1atm) occupies 22.41 Liters
∴ 20 moles NH₃(g) => 20moles/22.4Liters/mole
= 0.8928 Liters ≅ 0.9 Liter (1 sig. fig)