1.

The volume of 0.6 M NaOH required to neutralise 30 cm^(3) of 0.4 M HCl is

Answer»

`40cm^(3)`
`30CM^(3)`
`20CM^(3)`
`10cm^(3)`

SOLUTION :Normality=molarity`XX`basicity or acidity (for HCl)
`N_(2)=0.4xx1=0.4N` basicity=1 (for NaoH acidity=1)
`N_(1)=0.6xx1=0.6N,V_(1)=?,V_(2)=30cm^(3)`
From the equation, `N_(1)V_(1)=N_(2)V_(2)`
`0.6xxV_(1)=0.4xx30`
`V_(1)=(0.4xx30)/(0.6)=20cm^(3)`


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