Answer:
Add 0.70833 mL in 8.5 L of water we get 1 N HCl.
Explanation:
We know that Normality of HCL is [tex]$12 \mathrm{~N}$[/tex]
According to the formula
[tex]$$\begin{aligned}&\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2} \\&\mathrm{~N}_{1}=12 \mathrm{~N} \\&\mathrm{~V}_{1}=? \\&\mathrm{~N}_{2}=1 \mathrm{~N} \text { (Given) } \\&\mathrm{V}_{2}=8.5 \mathrm{ml}\end{aligned}$$[/tex]
Substituting the values in the above formula we get
[tex]$$\begin{aligned}&12 \times V_{1}=1 \times 8.5 \\&12 V_{1}=8.5 \mathrm{~mL}\\&V_{1}=\frac{8.5}{12}\mathrm{~mL}\\\\& V_{1}=0.70833\mathrm{~mL}\\\end{aligned}$$[/tex]