A) 2.0 nm
B) 0.1 nm
C) 2.0 pm
D) 1.0 pm
Correct Answer: C
Solution :
\[Moles=\frac{m{{v}_{ml}}}{1000}=\frac{{{10}^{-3}}\times 10}{1000}={{10}^{-5}}mole\] |
\[{{10}^{-5}}{{N}_{A}}\]molecules covering area\[=0.24\text{ }c{{m}^{2}}\] |
\[1{{N}_{A}}\]molecules covering area\[=\frac{0.24}{{{10}^{-5}}{{N}_{A}}}c{{m}^{2}}\] |
\[\frac{0.24}{{{10}^{-5}}\times 6\times {{10}^{23}}}={{a}^{2}}\] |
\[{{a}^{2}}=4\times {{10}^{-20}}c{{m}^{2}}\] |
\[a=2\times {{10}^{-10}}cm\] |
\[a=2\times {{10}^{-12}}m\] |
\[a=2pm.\] |
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