posted by anonymous .
Two aqueous solutions of sucrose, C12H22O12, are mixed. One solution is 0.1495 M sucrose and has d = 1.018 g/mL; the other is 12.00 % sucrose by mass and has d = 1.038 g/ml.
Calculate the moles C12H22O12 percent in the mixed solution.
Please provide steps
(assume volume is same for both solns)
How much of solution 1 is mixed with how much of solution 2. Won't it make a differen
First, the formula for sucrose is C12H22O11.
For the 0.1495 M soln.
density = 1.018 g/mL so 1 L has a mass of
1.018 g/mL x 1000 mL = 1018 grams.
How much sucrose is in that?
0.1495 moles x molar mass sucrose = 0.1495 x 342 (approximately, you look it up and do the calculations more accurately). = 51.1 grams C12H22O11.
mass water = 1018-51.1 = ?? g H2O.
For the 12.00% by mass solution.
density = 1.038 g/mL. Mass of 1L of this solution is 1.038 g/mL x 1000 mL = 1038 grams for the solution.
12% of that is sucrose; therefore
1038 x 0.12 = 124.6 g C12H22O11(again, I've rounded here and there).
mass water = 1038-51.6 = ?? g H2O
I don't think we can go any further until we know how much of each solution was mixed to form the final solution. Or have I missed something?