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If mass $%$ of oxygen in monovalent metal carbonate $48 %$ , then the number of atoms of metal present in $5 mg$ of this metal carbonate sample is : $(N_{A} = 6.0 \times 10^{23})$

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$30 \times 10^{16}$

$3 \times 10^{21}$

$6 \times 10^{19}$

$6.0 \times 10^{18}$

answer is B.

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Detailed Solution

$\begin{array}{l} M_{2} {CO}_{3} \\ \Rightarrow 48 = (\frac{48}{2 M + 60}) \times 100 \\ \Rightarrow 2 M + 60 = 100 \\ n = \frac{5 \times 10^{- 3}}{100} = 5 \times 10^{- 5} mol \\ n_{M} = 10 \times 10^{- 5} mol = 10^{- 4} mol \end{array}$

No. of atoms of $M = 6.00 \times 10^{19}$ atom

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