Answer:
125 ml of HCl
Explanation:
The molarity of the stock solution to determine how many milliliters would contain 1.5 moles of HCl. Since a concentration of 12.0 mol/L means that you get 12.0 moles of hydrochloric acid per liter of solution,
Concentration of required HCl (C1) = 1.0M
Volume of required HCL (V1) = 1500 ml
Concentration of stock HCl (C2) = 12M
Volume of stock HCL (V2) = ?
C1V1 = C2V2
V2 = C1V1/C2 = 1*1500/12 = 125 ml
