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Question

Chemistry Question on Some basic concepts of chemistry

Number of atoms of oxygen present in 10.6 g10.6\, g of Na2CO3Na _{2} CO _{3} will be

A

6.02×1023 6.02\times 10^{23}

B

12.04×1022 12.04\times 10^{22}

C

1.806×1023 1.806\times 10^{23}

D

31.80×102 31.80\times 10^{2}

Answer

1.806×1023 1.806\times 10^{23}

Explanation

Solution

Molecular mass of Na2CO3=2×23+12+3×16=106Na_2CO_3 = 2 \times 23 + 12 + 3 \times 16= 106 ∵106 g Na2CO3\because 106\, g\,Na_2CO_3 contains =3×6.023×1023= 3 \times 6.023 \times 10^{23} oxygen atoms ∴10.6 g\therefore 10.6\, g of Na2CO3Na_2CO_3 will contain =3×6.023×1023106×10.6= \frac{3 \times 6.023 \times 10^{23}}{106} \times 10.6 =18.069×1024= 18.069 \times 10^{24} =1.806×1023= 1.806 \times 10^{23} oxygen atoms