1.

The current in amp required to liberate Ag from a solution containing 1.7 xx 10^(-3) kg of AgNO_(3) was electrolysed in 1 hr is

Answer»

`1.342`
`0.27`
`0.027`
`0.1342`

Solution :1 MOLE of `AgNO_(3)-=` 1 Mole of Ag
`170 xx 10^(-3)` KG `AgNO_(3) -= 108 xx 10^(-3) ` kg of Ag
`therefore 1.7 xx 10^(-3)` kg `Ag NO_(3) = W_(Ag)`
`therefore W _(Ag) = (108 xx 10^(-3) xx 1.7 xx 10^(-3))/(170 xx 10^(-3))`
`= 1.08 xx 10^(-3)` kg
According to Faraday's `II^(ND)` law
`i = (W xx F)/(t xx E) = (1.08 xx 10^(-3) xx 96500)/(1 xx 60 xx 60 xx 108 xx 10^(-3)) = 0.27` amp


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