# Molarity/molality

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A 3.29 m solution of the solute H2O dissolved in the solvent acetone is available. Calculate the mass ( kg ) of the solution that must be taken to obtain 26.70 mol of H2O.

m= molality

I have no idea how to approach this problem, I'm not interesting in getting the answer, I actually what to learn how to do it.
Thanks =D

• Molarity/molality - ,

molality= molessolute/kgsolvent

3.29=26.70molH20/kg solvent

solve for kg solvent. Then add the mass equivalent of 26.70mole water to get kg of solution.

• Molarity/molality - ,

I found another post on this site, which explains a similar problem.
:)

• Molarity/molality - ,

Thank you!

• Molarity/molality - ,

Here is what I would do but check my thinking.
26.70 moles = 26.70 x 18.015 g H2O/mol = 59.269 g H2O (I know that's too many significant figures but I keep them in my calculator and round at the end).
So the SOLUTION has a mass of 59.269 g + 1000 g = 1059.269 g. Thus, 1000 g of that solution contains how many g H2O?
59.269 g H2O x (1000/1059.269) = 55.953 g H2O which is 3.105 moles in the 1000 g SOLUTION. We want 26.7 moles.
1000 g x (26.70 moles/3.105 moles) = 8599.03 g solution.
Now we must determine how much of that 8599.03 g is solute and how much solvent. That should do it.
I like to take this solution backwards now and see if it is 3.29 m. SOOO,
26.70 moles H2O x (18.015 g/mol) = 481 g H2O. Total solution is 8599.03 and -481 = 8118.03 g acetone + 481 g H2O
Therefore, m = moles/kg solvent = 26.70/8.118.03 = 3.289 which I would round to 3.29. Again, check my thinking.

• Molarity/molality - ,

This makes more sense, but your first step is a little confusing. when I do 26.70 x 18.015 g H2O/mol I get 481.005

but when I do:
3.29 m x 18.015 g H2O/mol then I get your ans: 50.269 g H20

• Molarity/molality - ,

It bothered me that the answer by Bob Pursley didn't agree better with my answer. The reason is that the calculation I did of
1000g x (26.70/3.1059) = 8596.54 but I typed in 3.105 which is the difference.

• Molarity/molality - ,

I just dropped the last 5.

• Molarity/molality - ,

And I'm a little unsure as to why after you find the mass of the solution 1059.269 you find how many g of H20 are in 1000 g of that soln. Why can't we just use the 1058.269 g? Was it to get the perfect 1000g (1kg) or was that unnecessary?

• Molarity/molality - ,

In regards to your latest post, I meant the very first step, not the step you did to check the answer. I'm assuming the first line is a typo?

• Molarity/molality - ,

I'm not sure of the question but if I interpret it correctly, you would simply subtract 1059.269-59.269 to get 1000 g H2O but we know that. What the problem asks us to do is to take 1000 g of the SOLUTION which will actually contain 55.953 g H2O (not 59.269g) and the difference (944.047 g)acetone solvent (not 1000 g).

• Molarity/molality - ,

Ohhhhh
That makes so much more sense. I didn't fully understand what the problem was asking.

Thank you so very much =)

• Molarity/molality - ,

I did make a typo in this last answer and I've corrected it below.
I'm not sure of the question but if I interpret it correctly, you would simply subtract 1059.269-59.269 to get 1000 g H2O(Should have said 1000 g acetone) but we know that. What the problem asks us to do is to take 1000 g of the SOLUTION which will actually contain 55.953 g H2O (not 59.269g) and the difference (944.047 g)acetone solvent (not 1000 g).