# Calculate % by mass in solution

Submitted by Wesley117 on Sun, 12/05/2010 - 14:32

A student is making 250.00 ml of a 0.10 M C6H12O6 (sugar)aqueous solution.  The

1. Calculate the mass of solute to be added to the solvent.
For this I came up with 4.505 g solute. (.250 x .1 x 180.2)

2. Calculate % by mass of sugar in the solution.
For this I would need mass of solute (4.505g)/mass of solute(4.505g) + mass of solvent.

How do I determine the mass of the solvent?

If the concentration is truly 0.10 M (0.10 molar), you'd need more data to be really accurate.
0.10 M = 0.10 mol/1.0L
(If the concentration were given in molality rather than molarity, as 0.10 m, you would have all data needed, because 0.10 m = 0.10 mol/1000 g solvent).

The combined mass of solute plus solvent for a 0.10 M solution is not exactly known from your data. (the temperature is not specified either, so I would assume 20oC). You'd need the specific volume or density for the solution, and I do not know where to find the data. The density will be close to 1, but not exactly 1.000 g/mL. Given your concentration (about 18 g/L or 18 mg/mL), assuming a density of 1.00 g/mL could be accurate enough. I work with solutions that are about 100 mg/mL in sugars and have estimated their densities as 1.03 g/mL.
For a solution density of 1.00 g/mL, 250.00 mL of solution would have a total mass (solute plus solvent) of 250 g.