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On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chlorine liberated is (At. Mass of `Cl=35.45`)A. 35.45gB. 70.9gC. 3.545gD. 17.77g |
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Answer» Correct Answer - C `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 . |
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