A) \[1.67\times {{10}^{21}}{{H}_{2}}O\]molecules
B) \[1.67\times {{10}^{26}}{{H}_{2}}O\]molecules
C) \[1.806\times {{10}^{23}}{{H}_{2}}O\]molecules
D) \[1.806\times {{10}^{21}}{{H}_{2}}O\]molecules
Correct Answer: A
Solution :
\[d=\frac{m}{v}\] or mass of 0.05 mL water \[=0.05\,mL\times 1.0\,g\,m{{L}^{-1}}\] \[=0.05\,g\,{{H}_{2}}O\] Moles of \[{{H}_{2}}O=\frac{0.05}{18}mol\,{{H}_{2}}O\] \[=\frac{6.023\times {{10}^{23}}\times 0.05}{18}\] \[=1.67\times {{10}^{21}}\,{{H}_{2}}O\,\text{molecules}\]You need to login to perform this action.
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