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September 25, 2016
Posted by **ChEm** on Friday, May 7, 2010 at 9:26pm.

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 -
**bobpursley**, Friday, May 7, 2010 at 9:41pmmolality= 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 -
**ChEm**, Friday, May 7, 2010 at 9:42pmI found another post on this site, which explains a similar problem.

:) - Molarity/molality -
**ChEm**, Friday, May 7, 2010 at 9:44pmThank you!

I got the correct answer. - Molarity/molality -
**DrBob222**, Friday, May 7, 2010 at 10:02pmHere 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 -
**ChEm**, Friday, May 7, 2010 at 10:17pmThis 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 -
**DrBob222**, Friday, May 7, 2010 at 10:18pmIt 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 -
**DrBob222**, Friday, May 7, 2010 at 10:21pmI just dropped the last 5.

- Molarity/molality -
**ChEm**, Friday, May 7, 2010 at 10:23pmAnd 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 -
**ChEm**, Friday, May 7, 2010 at 10:29pmIn 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 -
**DrBob222**, Friday, May 7, 2010 at 10:39pmI'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 -
**ChEm**, Friday, May 7, 2010 at 10:45pmOhhhhh

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

Thank you so very much =) - Molarity/molality -
**DrBob222**, Friday, May 7, 2010 at 11:20pm**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).