A solution is made by mixing 20.0 mL of toluene C6H5CH3d=0867gmL with 150.0 mL of benzene C6H6d=0874gmL. Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are

Question

A solution is made by mixing 20.0 mL of toluene C6H5CH3d=0867gmL with 150.0 mL of benzene C6H6d=0874gmL. Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are

____M and

_____m.

Answer 1

molarity=molestolulent/volume solution

volume solution=.150+.020
Moles toulene=20*.0867g/ml*molmass

molarity=molestoluene/kg solvent

kgsolvent=150*.0874 kg

I am wondering what was the problem on this?