A) \[6.02\times {{10}^{20}}\]
B) \[6.02\times {{10}^{16}}\]
C) \[6.02\times {{10}^{22}}\]
D) \[6.02\times {{10}^{21}}\]
Correct Answer: A
Solution :
Normality = Molarity\[\times \]Valency factor (n)\[0.02=M\times 2\] \[\text{Molarity}=\frac{\text{Number}\,\text{of}\,\text{moles}}{\text{Volume}\,\text{in}\,\text{litre}}\] \[0.01=\frac{n}{0.1}=n=0.001.\] mole of\[{{H}_{2}}S{{O}_{4}}\]contains\[6.023\times {{10}^{23}}\] molecules So, 0.001 mole contains \[6.023\times {{10}^{20}}\]molecules. Hence, the correct option is [a].
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