1.

Which amount of chlorine gas liberated at anode, if 1 ampere current is passed for 30 minutes from NaCl solution?

Answer»

`0.66` moles
`0.33` moles
`0.66` g
`0.33`g

Solution :`2CL^(-) to Cl_2 + 2e^(-)`
`Q = It "" ` Amonut of current passed `= 1 xx 30 xx 60 = 1800 C`
The amount of `Cl_2` liberated by passing 1800 coulomb of electric charge
`=(1 xx 1800 xx 71)/(2 xx 96500)`
`= 0.66 g`.


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