Molarity question????

Question

Calculate the molarity and mole fraction of acetone in a 1 m solution of acetone (CH3COCH3) in ethanol (C2H5OH). The density of acetone is 0.788g/cm3 and ethanol is 0.789 g/cm3. assume that the volumes of acetone and ethanol add.
ahh i dont kno how to approach this!!!!!!

the 1 m is molality, not molarity, lol sorry for the confusion.

Answer 1

molality is mole of solute per Kg of solvent
MW of acetone is 58.
So 1m =58 g acetone/ 1000 g ethanol

d=mass/V => V=mass/d
For acetone V =58/0.788 =73.60 ml
For ethanol V =1000/0.789 =1267.43 ml
So the total volume is =73.60+1267.43 =1341.03 ml = 1.34103 L of solution. In this volume we have 1 mole of acetone. Thus

M=mole/V(solution)= 1/1.34103 = 0.746 M

1000 g of ethanol correspond to 1000/46= 21.74 mole
The solution is 1m =>
1 mole acetone per 1000 g ethanol =>
1 mole acetone per 21.74 mole ethanol =>
1 mole acetone in total 21.74+1=22.74 mole =>
f(acetone)=1/22.74= 0.044

Answer 2

Well, you gave the molarity --> 1M. And keep in mind, 1 cm3 = 1 mL.

We can calculate the mole fraction as follows.

First, the volume of 1 mol of acetone can be calculated as:

1 mol x 58.079 g/mol / (0.788 g/mL) = 73.7 mL.

The rest of the volume of 1 L will be made up by the ethanol. So volume EtOH = 1000 mL - 73.7 mL = 926 mL. That much EtOH can be converted to moles EtOH:

926 mL x 0.789 g/mL / (46.07 g/mol) = 15.9 mol EtOH

So the mol ratio of acetone to EtOH is 1/15.9.