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A 35% solution of LiCl was electrolyzed ...

A `35%` solution of `LiCl` was electrolyzed by using a `2.5 A` current for `0.8 h. ` Assuming the current efficiency of `90%` , find the mass of `LiOH` produced at the end of electrolysis. `(` Atomic mass of `Li=7)`

Text Solution

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Number of Faradays `=(90)/(100)((2.5xx0.8xx3600)/(96500))`
`=0.067F`
`Ew=Mw` of `LiOH=7+16+1=24g mol^(-1)`
`(n` factor `=1)`
`W_(LiOH)=EwxxEq`
`=24 xx 0.067=1.61g`
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