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Volume of 0.6 M NaOH required to neutra...

Volume of `0.6` M NaOH required to neutralise `30cm^(3)" of "0.4M` HCI is

A

`40cm^(3)`

B

`30cm^(3)`

C

`20cm^(3)`

D

`10cm^(3)`

Text Solution

Verified by Experts

The correct Answer is:
C
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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