chemistry
posted by Dora on .
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

Use density to determine mass toluene, then convert mass to mols toluene.
Use density to determine mass 150.0 mL benzene,
Then molality toluene = #mols/kg solvent.
molarity = mols/L solution.
Post your work if you get stuck. 
i still got it wrong
0.867 g/mol= x/ 20 ml x= 17.94 g
0.874 g/mol= x/ 150 mil x= 131.1 g
toluene 17.904 g x 1 m/ 92.14 g/ mol= .194 m
benzee 131.1 g x 1m/78.1121 g/mol = 1.68 m
.194 m Toluene/ .1311 kg = 1.48 
0.867 g/mol= x/ 20 ml x= 17.94 gYou made a math error here. I get 17.34 g
0.874 g/mol= x/ 150 mil x= 131.1 gThis is OK.
toluene 17.904 g x 1 m/ 92.14 g/ mol= .194 m This should be 17.34 g toluene. And that divided by 92.14 = 0.188 mols toluene. Then 0.188/kg solvent = 0.188/0.1311 = 1.434 which I would round to 1.43 for molality toluene.
benzee 131.1 g x 1m/78.1121 g/mol = 1.68 m
.194 m Toluene/ .1311 kg = 1.48 see above for correction.
For molarity of toluene, we have 0.188 mols/150 mL or 0.188/0.150 L = 1.25 mols/L = 1.25 M.