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Posted by on Thursday, November 1, 2012 at 4:18pm.

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)

  • chemistry - , Thursday, November 1, 2012 at 6:34pm

    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

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