The volume of $10 \ N$ and $4 \ N$ $HCl$ required to make $1 \ L$ of $7 \ N$ $HCl$ are
- A$0.50 \ L$ of $10 \ N$ $HCl$ and $0.50 \ L$ of $4 \ N$ $HCl$
- B$0.60 \ L$ of $10 \ N$ $HCl$ and $0.40 \ L$ of $4 \ N$ $HCl$
- C$0.80 \ L$ of $10 \ N$ $HCl$ and $0.20 \ L$ of $4 \ N$ $HCl$
- D$0.75 \ L$ of $10 \ N$ $HCl$ and $0.25 \ L$ of $4 \ N$ $HCl$
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