100 cc of `0.6 N H_(2)SO_(4)` and 200 cc of `0.3 N HCI` were mixed together. The normality of the solution will be

Options

A. 0.2N
B. 0.4N
C. 0.8N
D. 0.6N

Answer

B

Explanation

Total volume of solution `= 100+200=300` cc
For `H_(2)SO_(4)`
`N_(1)V_(1)=N_(2)V_(2)`
`0.6N xx100=N_(2)xx300`
`N_(2)=(60N)/(300)=(N)/(5)`
For HCl
`N_(1)V_(1)=N_(2)V_(2)`
`0.3N xx200=N_(2)xx300`
`N_(2)=(N)/(5)`
Total normality `= N(H_(2)SO_(4))+N(HCl)`
`=(N)/(5)+(N)/(5)=(2N)/(5)=0.4N`