posted by Riley on .
A solution is made by mixing 25.0 mL of toluene C6H5CH3 d=0867gmL with 135.0 mL of benzene C6H6 d=0874gmL . Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are
how would i start this problem?
Start by convert 20.0 mL toluene to mass, then convert that to mols. mols = g/molar mass.
Then convert benzene to mass.
molality toluene = mols toluene/kg benzene.