A) \[6.022\times {{10}^{22}}\]
B) \[6.02\times {{10}^{21}}\]
C) \[6.02\times {{10}^{20}}\]
D) \[6.02\times {{10}^{18}}\]
Correct Answer: C
Solution :
For\[{{H}_{2}}S{{O}_{4}}\], Normality, N = 2M or \[M=\frac{N}{2}=\frac{0.02}{2}=0.01\] Moles of \[{{H}_{2}}S{{O}_{4}}=\frac{0.01\times 100}{1000}\] \[=1\times {{10}^{-3}}\] \[\therefore \] Number of molecules in 100 mL of 0.02N \[{{H}_{2}}S{{O}_{4}}=1\times {{10}^{-3}}\times {{N}_{A}}\] \[=1\times {{10}^{-3}}\times 6.023\times {{10}^{23}}\] \[=6.023\times {{10}^{20}}\]
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