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)

Answer»

`2.016" L "`
1.008L
11.2 L
20.16 L

Solution :(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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