How many grams of KNO3 will be required to prepare 800 mL of 1.4 M KNO3?

mass KNO3 = 113.236 g

Explanation:

∴V sln = 800 mL = 0.800 L

∴ C KNO3sln = 1.4 M

mass KNO3:

∴ molar mass KNO3 = 101.1032 g/mol

⇒ mol KNO3 = (0.800 L)*(1.4 mol/L) = 1.12 mol

⇒ mass KNO3 = (1.12 mol)*(101.1032 g/mol) = 113.236 g

i just learnt this recently but i need to look through my notes again