Number of electrons lost dring electrolysis of `0.355g` of `Cl^(c-)` is `(N_(A)=`Avogadro's number `)`

Options

A. `0.01`
B. `0.01N_(A)`
C. `0.02N_(A)`
D. `(0.01)/(2N_(A))`

Answer

b

Explanation

`Cl^(c-)+e^(-)rarr (1)/(2) Cl_(2)`
`1F=1e^(-) = 1 mol of e^(-) = N_(A)=35.5g Cl_(2)`
`:. 35.5 g of Cl_(2)=(N_(A)xx0.355)/(35.5)`
`=0.01 N_(A)`