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InterviewSolution
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The volume of 0.6 M NaOH required to neutralise 30 `cm^(3)` of 0.4 M HCl isA. `40cm^(3)`B. `30cm^(3)`C. `20cm^(3)`D. `10cm^(3)` |
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Answer» Correct Answer - 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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