1.

An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M acetic acid required to neutralise NaOH obtained above is

Answer»

`2000CM^(3)`
`100CM^(3)`
`200cm^(3)`
`1000cm^(3)`

Solution :WEIGHT of `NaCl=6.5xx(90)/(0.1)=5.85g`
No. of equivalents of NaCl
`=("wt. of NaCl")/("Mol. Mass of NaCl")=(5.85)/(58.5)=0.1`
No. of equivalents of NaOH obtained=0.1
`thereforeN_(NaOH)xxV_(NaOH)=N_(CH_(3)COOH)xxV_(CH_(3)COOH)`
`implies0.1xx1000=1xxV_(CH_(3)COOH)`
`impliesV_(CH_(3)COOH)=100cm^(3)`


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