Asked by ted
i am stuck on this problem from school
A solution is made by mixing 30.0 mL of toluene C6H5CH3d=0.867gmL with 130.0 mL of benzene C6H6d=0.874gmL . Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are
A solution is made by mixing 30.0 mL of toluene C6H5CH3d=0.867gmL with 130.0 mL of benzene C6H6d=0.874gmL . Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are
Answers
Answered by
bobpursley
How many grams of toluene do you have? How many moles is that?
What is the mass of the benzene?
Molarity=molesToulene/volumesolutioninLiters
molality=molesToulene/kgOfSolvent
What is the mass of the benzene?
Molarity=molesToulene/volumesolutioninLiters
molality=molesToulene/kgOfSolvent
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