1.

Value of V_(NaOH) is

Answer»

`291ml`
`145.5ml`
`75ML`
`582ml`

Solution :`M_(H_(2)SO_(4))=24.7/98xx(1000xx1.18)/100=2.97`
`N_(H_(2)SO_(4))=5.94`
Meq. of `H_(2)SO_(4)="meq."` of `NAOH+"meq. of" Al`
`5.94xx75=0.5xxV_(NaOH)+300`
`V_(NaOH)=291ml`


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