1.

A certain quantity of electricity is passed through aq. Al_(2)(SO_(4))_(3) and CuSO_(4) solutions connected in series 0.09 g of Al is deposited on cathode during electrolysis . The amount of copper deposited on cathode in grams is (At., mass of Al = 27 , Cu = 63.6)

Answer»

`3.18`
`0.318`
`31.8`
`0.636`

Solution :A/c. to Faraday's SECOND LAW
`(W_(Al))/(W_(Cu)) = (E_(Al))/(E_(Cu))`
`therefore W_(Cu) = (W_(Al) + E_(Cu))/(E_(Al)) = (0.09 XX 63.6 // 2)/(9) = 0.318 g`


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