1.

How much chlorine is obtained when 0.5 Faraday current is pass through aqueous solution of NaCl ? (Atomic weight of Cl=35.5 gm/mol)

Answer»

71.0 GM
35.5 gm
142.0 gm
17.75 gm

Solution :`NACL to Na^(+)+CL^(-)`
`Cl^(-) to (1)/(2) Cl_(2)+E^(-)0`
`(1)/(2)` mole gives 1 F
So, 1 F gives `(1)/(2)` mole=35.5 gm `Cl_(2)`
So, 0.5 F gives `(0.5xx35.5)/(1)=17.75` gm `Cl_(2)`.


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