A) \[6.023\times {{10}^{19}}\]
B) \[1.084\times {{10}^{18}}\]
C) \[4.84\times {{10}^{17}}\]
D) \[6.023\times {{10}^{23}}\]
Correct Answer: A
Solution :
\[0.0018 ml {{H}_{2}}O = 0.0018 g {{H}_{2}}O\] \[\because Density, d = 1 gm{{l}^{-1}}\] \[\therefore 1 mol {{H}_{2}}O = 18g\] \[\operatorname{i}.e = 6.023\times 1{{0}^{23}} moleudes\]
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