On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chlorine liberated is (At. Mass of `Cl=35.45`)

Options

A. 35.45g
B. 70.9g
C. 3.545g
D. 17.77g

Answer

C

Explanation

`Cl^(-)rarr(1)/(2)Cl_(2)+e^(-)`
I .e., 1 Faraday liberates `Cl_(2)=(1)/(2)"mol"=35.45g`.
`therefore` 0.1 Faraday will liberate `Cl_(2)=3.545` g .