a) You use 100 ml ethylchloride (density 0.92 g/ml) to extract 25 ml ethylene bromohydrin (density 2.41 g/ml) mixed with 100 ml. Is the aqueous layer the lower or the upper layer. How could you easily test this.

----- Ethylchloride is less dense so it is the top layer
Ethylene Bromohydrin is more dense so it is the bottom layer

b) You dissolve sodium chloride 40g. in the above mixture. Assuming the aqueous layer will have a volume of 112 ml, will the aqueous phase be the lower or the upper layer? Explain.

Can you check the first answer and help me with the second one?

Thanks!

1 answer

You have posted this for the last day or two. One reason you don't have an answer is that you don't say what the ethylene bromohydrin is mixed with. You say mixed with 100 mL OF WHAT? Second reason is when I look up ethylene bromohydrin on google I find the density is 1.79 g/mL and not 2.41.

The b part isn't clear to me what the mixture is.

If the a part is 25 mL ethylene bromohydrin in 100 mL H2O, I would assume (and the assumption probably isn't correct but I think that's all you can do) that the volumes are additive.
Then grams 100 mL H2O if the density is 1.00 g/mL = 100 grams.
grams 25 mL ethylene bromohydrin = density (whatever that is) x 25 mL = ?
Total volume = 100 + 25 = 125 mL
Total mass = 100 + mass bromohydrin
Then density = mass/volume = ?/125 = ?
If I do this with the density you list in the post as 2.41 I come up with something like 1.28 g/mL which makes the density > 0.92 so the aqueous layer would be on the bottom. Check my thinking. I think your answer assumes that the ethylene bromohydrin/water mix has the same density as the pure bromohydrin and I don't think that's a good assumption; however, you answer may still be right if my calculation stands up. I also calculated it as above based on 1.79 g/mL density and my calculation came up with about 1.15 g/mL for the mix which still is > 0.92.
I don't know what to do with part b since I don't know what the statement means.