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
100 mL for each soln.
density = 1.018 g/mL x 100 mL = 101.8 g = mass solution.
moles = M x L = 0.1495 x 0.100 = 0.01495
g sucrose = 0.01495 x 342(molar mass but you look it up because I've rounded here and there) = 5.11 g sucrose which means the water has a mass of 101.8-5.11 = ?? g H2O.
density = 1.038 g/mL x 100 mL = 103.8 g = mass solution
12.00% sucrose or 103.8 x 0.1200 = 12.46 g sucrose or 12.46/342 = 0.0364 moles sucrose.
Water has a mass of 103.8-12.46 g = ?? g.
Now take moles from A and add to moles B to obtain total moles sucrose.
Add grams water from A to grams water from B to obtain total moles H2O.
Find mole fraction sucrose (mols sucrose/(moles sucrose + moles H2O) = Xsucrose
Then mole percent = 100*Xsucrose