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 $N{a}_{2}C{O}_3=2\times 23+12+3\times 16=106$ $\because 106gN{a}_{2}C{O}_3$ 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