posted by jack .
A 15.0ml sample of NaCL solution has a mass of 15.78g.after the NaCL solution is evaporated to dryness, the dry salt residue has a mass of 3.26g. calculate the following concentrations for the %(m/m), % (m/v), molarity (M)
The problem is stated a screwy way. It almost sounds as if it weighed 15.78 AFTER evaporation. I think the problem is two run on sentences.
15.78 = mass NaCl + H2O
3.26 = mass NaCl
12.52 = mass H2O
% m/m = (mass NaCl/total mass)*100 =
You could do molality = mols/kg solvent.
%m/v = (mass NaCl/volume)*100 =
v = 15.0 mL
M = mols/L soln. mol = gNaCl/molar mass
kg solvent = 0.01252