Hydrochloric acid is sold commercially as 12M solution. How many moles and how many grams of HCl are present in 300 mL of 12.0 M solution ?

Answer

3.6 mol, 131.4 g of HCl

Explanation

Molarity of solution `= ("Moles of HCl")/("Volume of solution in litres")`
`("12 mol L"^(-1))=("Moles of HCl")/("0.3 L")`
Moles of HCl `= ("12 mol L"^(-1))xx("0.3 L")=3.6` mol
Mass of HCl `=("3.6 mol")xx("36 g mol"^(-1))=131.4 "g HCl"`.