Volume of `3.6` M NaOH required to neutralise `30cm^(3)" of "0.4M` HCI is
Volume of `3.6` M NaOH required to neutralise `30cm^(3)" of "0.4M` HCI is
A. `20cm^(3)`
B. `40cm^(3)`
C. `45cm^(3)`
D. `30cm^(3)`
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Correct Answer - A
`M_(1)V_(1)=M_(2)V_(2)`
`0.6xxV_(1)=0.4xx30,V_(1)=(0.4xx30)/(0.6)=20cm^(3)`
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