Describe how you would prepare an aqueous solution of acetone (CH3COCH3) in which the mole fraction of acetone

Describe how you would prepare an aqueous solution of acetone (CH3COCH3) in which the mole fraction of acetone is 0.25.

The answer is each gram of acetone requires 0.93 g of water, but I have no idea how they came up with that answer, so if anyone can explain how to get that I would really appreciate it.

Answer:
if the mole fraction is 0.25 that means for every 0.25 mol acetone, then there would have to be 0.75 mol of water. (That is so that when you do moles acetone divided by total number of moles you get 0.25)

0.25 moles of acetone is 58g * 0.25 = 14.5 g

0.75 moles of water = 18g * 0.75 = 13.5g

divide the two by 14 and you should get a ratio of 1 to 0.93
hi here is the solution:

assume a basis of 100 moles of solution

thus moles of acetone= 25mol
moles of water =75mol

mass of acetone = 25mol x Mr of acetone
= 25 mol x 58g/mol
= 1450g
mass of water = 75mol x 18g/mol
=1350g
total mass = 2800g

mass fraction of acetone = 1450g/2800g =0.51785
mass fraction of acetone = 0.51785
mass of acetone/mass of solution = 0.51785
1g/mass of solution =0.51785
mass of solution =1.93g
mass of acetone +mass of water =1.93g
1g + mass of water =1.93g
thus mass of water =0.93g for every 1g acetone

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