posted by Anonymous on .
can someone help me with this please? 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.
I don't see a decimal in the density tabulations. That may be your first problem. The density of toluene is 0.867 that of benzene is 0.879 g/mL.
First, let's do some housekeeping.
mass toluene = 25.0 mL x 0.867 g/mL = 21.675 g
moles toluene = 21.675/92.14 = 0.235 moles.
mass benzene = 135.0 mL x 0.874 g/mL = 117.99 g
molality = moles solute/kg solvent
0.235/0.11799 = ??m, then round to 3 significant figures.
molarity = moles solute/L soln
0.235/(0.0250 + 0.1350) = 0.235/?? = xxM, then round to three s.f.
Check my work.