Question

A chemist wants to make 500.0 mL of 0.215M NaCl. What mass of NaCl is needed?

would it be
.500 L x .215 M = .1075 mol NaCl
.1075 mol x 58.44g NaCl
= 6.28g NaCl needed

Answer 1

Yes

Answer 2

Making solutions... Solutions are prepared form one of 3 sources, Manufactured Solid Stock, Manufactured Liquid Stock or Laboratory Concentrate.

For mass of solid mfg stock needed (grams) use formula...
gms = [(Molarity needed x Vol needed(in liters) x formula wt of solute)] / [decimal fraction Purity factor]

Form vol of liqd mfg stock (ml) = [(Molarity Needed)(Vol needed in liters)(formula wt of solute)] / (fractional purity factor)(Specific Gravity)

From Laboratory Concentrate: Vol(ml) of concentrate needed => use 'dilution equation' and solve for Vol of concentrate.

(M X V)conc = (M x V)diluted

Transfer calculated quantity into mixing container and add solvent up to but not to exceed needed solution volume. Exception - For concentrated acids, add the liquid concentrate into a small quantity of water to minimize over heating and finish with solvent up to desired solution volume.