Answer:
[tex]1542.85\ \text{mL}[/tex]
Explanation:
[tex]V_1[/tex] = Volume of HCl = 60 mL
[tex]C_1[/tex] = Initial concentration of HCl = 9 M
[tex]C_2[/tex] = Final concetration of HCl = 0.35 M
[tex]V_2[/tex] = Volume to be diluted
We have the relation
[tex]\dfrac{C_1}{C_2}=\dfrac{V_2}{V_1}\\\Rightarrow \dfrac{9}{0.35}=\dfrac{V_2}{60}\\\Rightarrow V_2=\dfrac{9}{0.35}\times 60\\\Rightarrow V_2=1542.85\ \text{mL}[/tex]
The required volume is [tex]1542.85\ \text{mL}[/tex].
