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
2 answers
Yes
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.
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.
1 answer