Question

To what volume will you have to dilute 60.0 mL of a 9 M HCl solution to

Answer 1

Answer:

$1542.85\ \text{mL}$

Explanation:

$V_1$ = Volume of HCl = 60 mL

$C_1$ = Initial concentration of HCl = 9 M

$C_2$ = Final concetration of HCl = 0.35 M

$V_2$ = Volume to be diluted

We have the relation

$\dfrac{C_1}{C_2}=\dfrac{V_2}{V_1}\\\Rightarrow \dfrac{9}{0.35}=\dfrac{V_2}{60}\\\Rightarrow V_2=\dfrac{9}{0.35}\times 60\\\Rightarrow V_2=1542.85\ \text{mL}$

The required volume is $1542.85\ \text{mL}$.