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\[{{H}_{2}}O=\frac{0.05}{18}mol\,\,{{H}_{2}}O\] \[=\frac{6.023\times {{10}^{23}}\times 0.05}{18}\] \[=1.67\times {{10}^{21}}{{H}_{2}}O\]molecules
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