a solution (d=1.049g/ml) was prepared by mixing hexane (C6H14 / d= 0.659g/ml) with toluene (C7H8 / d= 0.867g/ml). Determine the% (v/v) of the solution knowing it is 4.39M toluene.

Answer 1

4.39 M toluene = mols toluene/L solution
mols toluene = 4.39 moles in 1 L or 1,049 grams solution.
molar mass hexane, C6H14, = 86
molar mass toluene, C7H8, = 92
grams toluene in 4.39 M = mols x molar mass = 4.39 x 92 = estimated 404 grams.
volume toluene = mass/density = estimated 404/0.867 = estd 466 mL.
The mass of 1 L of the solution is 1,049 grams. The toluene in that solution has a mass of 404 g. Therefore, mass hexane is 1049 total - 404 = 645 g hexane. Volume of hexane = mas/density = 645/0.659 = estd 979 mL. Assuming the volumes are additive the total volume is 979 + 466 = 1445 mL. So percent toluene = (volume toluene/total volume)*100 = ?
percent hexane = (volume hexane/total volume)*100 = ?
I didn't check my calculations. Sometimes I punch the wrong buttons on the calculator. You should double check everything. Check my thinking. Sometimes I make a typo on the numbers so check that too. Finally, I have rounded here and there and estimated here and there . Hope this helps.