A) \[0.5\times {{10}^{-19}}{{m}^{2}}\]
B) \[4\times {{10}^{-20}}{{m}^{2}}\]
C) \[0.5\times {{10}^{-10}}{{m}^{2}}\]
D) \[5\times {{10}^{-19}}{{m}^{2}}\]
Correct Answer: D
Solution :
[d]| Amount of \[C{{H}_{3}}COOH\] adsorbed on Ig of charcoal\[=(0.5-0.49mo1{{L}^{-1}})\]\[\times \frac{100mL}{1000mL/L}=1\times {{10}^{-3}}mol\] |
| So, no .of \[C{{H}_{3}}COOH\]molecules in \[1\times {{10}^{-3}}\]mol of acetic acid \[=6.02\times {{10}^{23}}mo{{l}^{-1}}\times {{10}^{-3}}mol=6.02\times {{10}^{20}}\] |
| Area occupied by one \[C{{H}_{3}}COOH\] molecule \[=\frac{3.01\times {{10}^{2}}{{m}^{2}}}{6.02\times {{10}^{20}}}=5\times {{10}^{-19}}{{m}^{2}}\] |
| Thus the surface area of the charcoal covered by each \[C{{H}_{3}}COOH\]molecule is \[5\times {{10}^{-19}}{{m}^{2}}\] |
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