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The no. of silver atoms presents in 90% pure silver wire weighing 10g is (at. wt. of Ag = 108)
A.$8 \times {10^{22}}$
B.$0.62 \times {10^{23}}$
C.$5 \times {10^{22}}$
D.$6.2 \times {10^{29}}$

Answer
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Hint: In order to approach the solution of the question one should use the concept of moles whose formula is given as-
$number{\text{ }}of{\text{ }}moles = \dfrac{{mass{\text{ }}of{\text{ }}substance}}{{mass{\text{ }}of{\text{ }}one{\text{ }}mole}}$
where mass of one mole is atomic mass of silver (Ag).

Complete answer:
According to the given information we have silver wire which is $90\%$ pure whose weight is 10g and we have to identify the total numbers of silver atoms present in this silver wire
So, in order to find the numbers of silver atoms present in the wire let’s use the formula of number of moles which is given as; $number{\text{ }}of{\text{ }}moles = \dfrac{{mass{\text{ }}of{\text{ }}substance}}{{mass{\text{ }}of{\text{ }}one{\text{ }}mole}}$ we know that mass of one mole will be the atomic mass of silver (Ag) i.e. 108
And the mass of substance = 90% of 10 g of silver wire = $\dfrac{{90}}{{100}} \times 10g = 9g$
Substituting the values in the formula of number of moles we get
Number of moles = $\dfrac{9}{{108}} = 0.083$
Now we know that 1 mole of silver contain $6.022 \times {10^{23}}$silver atoms
Therefore, number of silver atoms = $0.083 \times 6.022 \times {10^{23}}$
$ \Rightarrow $Number of silver atoms = $0.5 \times {10^{23}}$ or $5 \times {10^{22}}$
Therefore, $5 \times {10^{22}}$atoms of silver are present in the 10g of silver wire which is $90\%$ pure.

Hence, option C is the correct option.

Note:
In the above solution we used the concept of moles which is the unit measurement of a given substance in international system of units i.e. SI so moles are the part of a substance which contains exactly $6.022 \times {10^{23}}$ particles in each mole. So, as in the above solution we got the number of moles of silver we multiplied it by $6.022 \times {10^{23}}$ in order to obtain the number of silver atoms.