Q1. Calculate the number of atoms in

a) 0.5 mole atoms of nitrogen

b) 0.2 mole moleules of nitrogen

c) 3.2g of sulphur

d)3.4 g of H2S

Q2. how many molecules of water of hydration are present in 252 mg of oxalic acid?

1) (a) 1 mole of nitrogen atoms = 6.022 ×1023 atoms
0.5 moles of Nitrogen atoms = 6.022 ×1023 × 0.5  = 3.011×1023 atoms

(b) 1 mole of nitrogen molecules  = ​ 6.022 ×1023  molecules
0.2 moles of nitrogen molecules = 6.022 ×1023 × 0.2  = 1.2044×1023 molecules
As 1 molecule of nitrogen contains 2 atoms of nitrogen.
So, ​1.2044×1023 molecules of nitrogen will contain 1.2044×1023× 2 atoms
= 2.4088 ×1023 atoms

(c) 32 g of sulphur (Molar mass of Sulphur ) = 6.022 ×1023 atoms
3.2 g of sulphur  =3.2 ×6.022×102332
= 0.6022 ×1023 atoms

(d) 3.4 g of H2S
Molar mass of H2S = 34 g mol-1
34 g of H2S contains 6.022×1023 molecules
1 molecule of H2S contain 2 atom of hydrogen and 1 atom of sulphur.
34 g of H2S contain 2×6.022×1023 atoms of hydrogen and 6.022×1023 atoms of Sulphur.
3.4 g of H2S contain 12.044×1022 atoms of hydrogen and 6.022×1023 atoms of sulphur.
2) Molecular formula of oxalic acid is H2C2O4.2H2O
and molar mass of oxalic acid  = [2×1 +12×2+4×16 + 2×18] = 126 gmol-1
water molecule in 1 mole oxalic acid  = 2 molecules
So, 126 g of oxalic acid contains 2 × 6.022 ×1023 molecules of water. 
thus, 0.252 g of oxalic acid will contain = 0.252 ×2×6.022×1023126
=  2.409  ×1021 molecules of water 



 

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