Question

Since Avogadro number is $6.023\times {10}^{23}$ and atomic mass of nitrogen is $14$ amu, one mole of $N_2$ gas will have:

• A. $0.5\times 6.023\times {10}^{23}$ atoms
• B. $2\times 6.023\times {10}^{23}$ atoms
• C. $14\times 6.023\times {10}^{23}$ atoms
• D. $28\times 6.023\times {10}^{23}$ atoms

Answer 1

Answer: (B)

$2\times 6.023\times {10}^{23}$ atoms

1 molecule of $N_2$ = 2 atoms of $N$

1 mole of $N_2=2\times N_A$ atoms of $N$ (since, one mole$=6.023\times {10}^{23}$ number of entities)
1 mole of $N_2$ gas $=2\times 6.023\times {10}^{23}$ atoms of $N$.