Question

I had posted a while back and still couldn't solve this is there anyway you could show me the steps on how u did this? thank you very much

A solution is made by mixing 20.0 mL of toluene C6H5CH3d=0.867gmL with 150.0 mL of benzene C6H6d=0.874gmL.

Assuming that the volumes add upon mixing, what are the molarity (M) and molality (m) of the toluene

Answer 1

Dora--I scrolled back yesterday and noticed you had added some material but I also noticed a math error. I posted a correction but I don't know if you saw it or not. Here is the way you do the problem. Check my arithmetic very carefully.

mass 20.0 mL toluene = 20.0 x 0.867 = 17.34 grams (I think you had 17.9 something).
mols toluene = 17.34/92.14 = 0.188 mols toluene.

mass benzene = 150.0 mL x 0.874 = 131.1 grams = 0.1311 kg.
molality toluene = 0.188 mol/0.1311 kg solvent = 1.4355 m which I would round to 1.44 m to three s.f.

molarity toluene in benzene = mols toluene/total L of solution = 0.188 mols toluene/0.170 L solution (that's 150.0 mL + 20.0 mL = 170 mL) = 1.10588 which rounds to 1.11 to three s.f.
Check my thinking. Check my work.

Answer 2

thank you very much Dr. Bob