Question

Number of atoms of oxygen present in 10.6 g of $ N{{a}_{2}}C{{O}_{3}} $ will be

• A. $ 6.02\times {{10}^{23}} $
• B. $ 12.04\times {{10}^{22}} $
• C. $ 1.806\times {{10}^{23}} $
• D. $ 31.80\times {{10}^{28}} $

Answer 1

Answer: (C)

Molecular mass of $ N{{a}_{2}}C{{O}_{3}}=2\times 23+12+3\times 16=106 $ $ \because 106\,g\,N{{a}_{2}}C{{O}_{3}} $ contains $ =3\times 6.023\times {{10}^{23}} $ oxygen atoms $ \therefore $ 10.6 g of $ N{{a}_{2}}C{{O}_{3}} $ will contain $ =\frac{3\times 6.023\times {{10}^{23}}}{106}\times 10.6 $ $ =18.069\times {{10}^{24}} $ $ =1.806\times {{10}^{23}} $ oxygen atoms