1.

5600 ml of O_(2) is obtained by electrolysis at STP, then how much Ag is obtained by same electricity ? (atomic mass of Ag=108)

Answer»

`5.4gm`
`10.8gm`
`54.0gm`
`108.0gm`

SOLUTION :`n_(O_(2))=(5600)/(22400)=(1)/(4)` MOL `O_(2)`
`(W_(Ag))/(108)xx1=(W_(O_(2)))/(M_(O_(2)))xx4""(2H_(2)O to O_(2)+4H^(+)+4e^(-))`
`(W_(Ag))/(108)xx1=(W_(O_(2)))/(M_(O_(2)))xx4`
`(W_(Ag))/(108)=(1)/(4)xx4""therefore W_(Ag)=108gm`.


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