# chemistry

posted by 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

• chemistry - ,

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.

• chemistry - ,

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

• chemistry - ,

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.