Question

what quantity of NaCl is required for preparing 200ml of 0.9%solution​

Answer 1

For example, 0.9% NaCl contains 0.9 g NaCl per 100 ml of solution or 9 g NaCl/l. This can be converted to molarity by dividing by molecular weight: 0.9% NaCl = (9g/l)/(58.5g/mole) = 0.15 M NaCl. Mole: A mole (or mol) of a substance is simply Avogadro's number particles of that substance.