An aqueous solution of NaNO3 has a concentration of 0.363 molality. Its density is 1.0185 g mL ^ -1. Calculate the molar concentration of NaNO3 and the mass percent of NaNO3 in the solution. What is the mole fraction of NaNO3 in the solution?
answers should be
0.359 Molarity
3.00% mass (percent NaNO3)
6.49 x 10 ^ -3 (mole fraction)
Can anyone please explain how to do the problem? I appreciate any help thank you
2007-03-02 03:07:06 · 1 answers · asked by E.T.01 5 in Science & Mathematics ➔ Chemistry
molality is mass of solute/mass of solvent in grams.
Therefore
.363 = 1.0185/x
=
x = 2.8 gms of solvent.
Therefore the molar concentration =
volume of solute/volume of solvent (ML) =
1.0185/2.8
= 360 + or - 1.
percentage composition =
Mass of the specific solute/
the total mass *100
=
2.8 + or -2%.
mole fraction =
mass of solute/total mass of compound =
1.0815/3.5
= 6.49 * 10^-3
2007-03-02 03:21:20 · answer #1 · answered by akshayrangasai 2 · 0⤊ 0⤋