By passing 0.1 Faraday of electricity through fused sodium chloride, the amount of chlorine liberated is

When aqeous NaCl is eectrolysed , Ther reaction at anode is :

$$\begin{gathered} C{l}^{-}\to \frac{1}{2}C{l}_2+\hspace{0.17em}{e}^{-} \\ i mole electrons = 1 Faraday \\ 1 F\hspace{0.17em}of electricity liberates = \frac{1}{2} mol of C{l}_2=35.45g \\ therefore 0.1F \hspace{0.17em}liberates = 3.545g \end{gathered}$$