posted by Sam on .
Okay, so I have to find the percent of water in each solution, but it involves different reagents and I'm not sure how to figure it out. Here's one example:
Add all together -
4 mL water
2 mL of 0.01 M aqueous NaOH
2 mL acetone
2 mL 0.1 M alkyl halide in acetone
How do I calculate how much is water from the aqueous NaOH and how much acetone is in the alkyl halide?
I'm a little confused as to the question. If you want percent water (not counting the solutions) it is (4/10)*100 = ?
If you want to count the water in the aq NaOH too that is almost 2 mL (0.01M NaOH is only 0.4 g NaOH in a liter which will be 0.4 mg in 1 mL or 0.8 mg in 2 mL which is about 2000 mg H2O - 0.8 mg NaOH = 1999.2 mg H2O and for all practical purposes that is 2 mL. We could do it more accurately if we knew the density of the 0.01 M solution.)
That will make it 4mL + 2 mL = 6 mL out of a total of 10 mL so percent H2O is (6/10)*100 = ?
There is no acetone in the water so you can ignore those calculations.