Question

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 1M acetic acid required to neutralise NaOH obtained above is
A. `2000cm^(3)`
B. `100cm^(3)`
C. `200cm^(3)`
D. `1000cm^(3)`

Answer 1

Correct Answer - B
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)`