A) \[3.39\times {{10}^{22}}\]
B) \[6.023\times {{10}^{23}}\]
C) \[1.20\times {{10}^{23}}\]
D) \[2.46\times {{10}^{23}}\]
Correct Answer: A
Solution :
\[2C{{l}^{-}}\xrightarrow[{}]{{}}C{{l}_{2}}+2{{e}^{-}}\] \[2\times 35.5g\,C{{l}^{-}}\]loses = 2 moles of electrons \[1gC{{l}^{-}}\]loses\[=\frac{2}{2\times 35.5}\]moles of electrons \[2g\,C{{l}^{-}}\]closes\[=\left( \frac{2}{2\times 35.5} \right)\times 2\] \[=\frac{2}{35.5}\]mole of electrons \[\Rightarrow \]\[\frac{2\times 6.023\times {{10}^{23}}}{35.5}=3.39\times {{10}^{22}}\]electrons
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