Question

If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?

Answer 1

$M_{A}V_{A}=M_{B}V_{B}\\(80.0)(4.00)=V_{B}(12.4)\\V_{B}=\frac{(80.0)(4.00)}{12.4} \approx \boxed{25.8 \text{ mL}}$