What is the number of atoms of each element present in $6.3$ g of nitric acid $(H N O_{3})$ ? (Atomic Mass of $H = 1 u$ . Atomic mass of $N = 14 u$ , Atomic mass of $O = 16$ u )

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Solution

Moles of $H N O_{3}$ = 6.3/63 = 0.1 moles
HN $O_{3}$ has 1 atom of H , 1 atom of N , 3 atoms of O
No of atoms = moles of molecule × atomicity of atom × avogadro number
No of atoms of H $= 0.1 \times 1 \times 6.023 \times 10^{23}$ = $6.023 \times 10^{22}$
No of atoms of N $= 0.1 \times 1 \times 6.023 \times 10^{23}$ = $6.023 \times 10^{22}$
No of atoms of O $= 0.1 \times 3 \times 6.023 \times 10^{23}$ = $18.069 \times 10^{22}$

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