Question

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

Answer 1

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.

Answer 2

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

Answer 3

0.867 g/mol= x/ 20 ml x= 17.94 g<b>You made a math error here. I get 17.34 g</b>

0.874 g/mol= x/ 150 mil x= 131.1 g<b>This is OK. </b>

toluene 17.904 g x 1 m/ 92.14 g/ mol= .194 m <b>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.</b>

benzee 131.1 g x 1m/78.1121 g/mol = 1.68 m

.194 m Toluene/ .1311 kg = 1.48 <b>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.</b>