The number of atoms of oxygen present in 10.6 g of $N a_{2} C O_{3}$ will be:

6.02 \times 10^{22}

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12.04 \times 10^{22}

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1.806 \times 10^{23}

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31.8

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Solution

The correct option is C $1.806 \times 10^{23}$
Molar mass of $N a_{2} C O_{3}$ $= (2 \times a t o m i c m a s s o f n i t r o g e n) + (1 \times a t o m i c m a s s o f c a r b o n) + (3 \times a t o m i c m a s s o f o x y g e n)$ $= 2 (23) + 1 (12) + 3 (16)$ $= 106 g / m o l$

So, $1$ mole of ${N a}_{2} {C O}_{3} = 108 g$

Moles in $10.6 g$ of ${N a}_{2} {C O}_{3} = 10.6 / 106 = 0.1$ mole

$106 g$ contain atoms of oxygen $= 3 \times 6.022 \times 10^{23}$

$10.6$ contain atoms $= 0.1 \times 3 \times 6.022 \times 10^{23}$
$= 1.8 \times 10^{23} a t o m s$

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