A current of 9.65 amperes is passed through excess of fused `AlCl_(3)` for 5 hours. How many litres of chlorine will be liberated at S.T.P. ? (1F=96500C)A. `2.016" L "`B. 1.008LC. 11.2 LD. 20.16 L

A current of 9.65 amperes is passed through excess of fused `AlCl_(3)`

1.	A current of 9.65 amperes is passed through excess of fused `AlCl_(3)` for 5 hours. How many litres of chlorine will be liberated at S.T.P. ? (1F=96500C)A. `2.016" L "`B. 1.008LC. 11.2 LD. 20.16 L
Answer» Correct Answer - D (d) The anodic reaction is : `2Cl^(-) to underset(22.4L)(Cl_(2))+underset(2F)(2e^(-))` The quantity of charge (Q) passed`=ixxt` No. of Faradays of `=(9.65xx5xx60(C ))/(96500(C ))` charge passed =1.8 F 1F of charge evolves `Cl_(2)=11.2 L` 1.8 F of charge evolves `Cl_(2)=(11.2L)/((1F))xx(1.8F)=20.16" L"`

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A current of 9.65 amperes is passed through excess of fused `AlCl_(3)` for 5 hours. How many litres of chlorine will be liberated at S.T.P. ? (1F=96500C)A. `2.016" L "`B. 1.008LC. 11.2 LD. 20.16 L

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