Question

Calculate the weight of following gases which occupy a volume 3.36 dm^3 at STP.

(l) Nitrogen

(ll) Hydrogen

(lll) Helium

Answer 1

$\bold{\text{number of moles}= \frac{\text{given volume}}{22.4L}}$
Here given volume = 3.6 dm³ = 3.36L { because 1 dm³ = 1 L }
now, number of moles of gas = 3.36/22.4 = 3/20 = 0.15

(a) Nitrogen ( N₂ )
molecular weight of N₂ gas = 28 g/mol
Number of moles = 0.15
so, weight of N₂ gas = molecular weight × number of moles
= 28 × 0.15 = 4.2 g

(b) Hydrogen ( H₂ )
Molecular weight of H₂ gas = 2 g/mol
so, weight of H₂ gas = molecular weight × number of moles
= 2 × 0.15 = 0.3 g

(c) Helium ( He)
Molecular weight of He gas = 4g/mol
so, weight of He gas = molecular weight × number of moles
= 4 × 0.15 = 0.6 g

Answer 2

Given volume is 3.36 dm^3=3.36L
1 mole of any gas occupies 22.4L.
1 mole ---------> 22.4L
? -------------------->3.36L
Number of moles = 3.36/22.4
=3360/22400
=0.15 mole

A) weight of Nitrogen-N2
Molecular mass if N2=28u
Molar mass of N2=28g
Number of moles= 0.15
Weight of N2 gas=number of moles x molecular weight
=0.15×28
=4.2g

B) weight of hydrogen -H2
Molecular weight of H2=2u
Molar mass= 2g
Weight of H2= Number molesx molecular weight
=0.15×2=0.3g

C) weight of Helium:
Molecular weight=4u
Molar mass=4g
Weight of He=4×0.15=0.6g