The volumes of $4\,N\, HCl$ and $10\,N\, HCl$ required to make $1\, L$ of $6\,N\, HCl$ are
- A$0.75\, L$ of $10\,N\, HCl$ and $0.25\, L$ of $4\,N\, HCl$
- B$0.25\, L$ of $4\,N\, HCl$ and $0.75\, L$ of $10\,N\, HCl$
- C$0.67\, L$ of $4\,N\, HCl$ and $0.33\, L$ of $10\,N\, HCl$
- D$0.80\, L$ of $4\,N\, HCl$ and $0.20\, L$ of $10\,N\, HCl$
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