posted by Anonymous on .
can someone please help me with this? i've been stuck on this one forever...
A solution is made by mixing 25.0 mL of toluene C6H5CH3d=0867gmL with 135.0 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.