Question

If I had 10 mL of water and dissolved 1.0 grams of salt, what would be the percentage of the water? The volume of the mixture?

Answer 1

1g of water = 1mL of water


thus we have

10g of water
and
1g of salt

Answer 2

do you know how to solve from here?

Answer 3

Um, no. That's pretty much why I asked the question :P I don't know how to get the percentage of water or the volume of the mixture.

Answer 4

Well derp, also this isn't a mixture it is a solution.

the percentage of water

( Mass of Water/(Mass of Water + Mass of Salt) )*100

so,

(10g/11g)*100 = 91%

As for determining the volume I have no idea how to calculate the displacement the NaCl would have on the 10mL of water you can just assume it is negligible and state it is 10mL but that is perhaps an incorrect assumption. I will try to google some more and see if someone has constructed a way of calculating displacement of NaCl in water but thats about as much as I can do.

Answer 5

Well, thanks anyway,

Answer 6

You probably would just have to measure the volume of the solution in a laboratory I'm unable to find any information on calculating displacement of NaCl in water might want to take this up with your teacher

Answer 7

The paper is due tomorrow since we did the lab today. It's just a matter of filling in the numbers. I will ask to see if he can give me some help.

Answer 8

A large table here (PDF):

http://seagrant.oregonstate.edu/sgpubs/onlinepubs/h99002.pdf

suggests that at 9% by weight there are 0.97 mL of pure water per mL of salt solution, so if you started with 10.0 mL of pure water, you'd end up with 10.3 mLof salt solution.

That's a pretty sophisticated calculation, however, and I'd be surprised if you were asked to do such a thing short of junior lab in college. Are you sure you need this information? Perhaps there is another way you're expected to calculate your results?

Answer 9

Nice find Carl Pham

Answer 10

Woah, I'm just a Sophomore in High school. Hah. My teacher would most likely show us how to get the volume of it if it was really difficult or something we haven't learned yet. I tried looking around for some kind of formula but I couldn't find anything.

Answer 11

The table in Carl_Pham post gives a table that provides a specific gravity 1.066 at 9% mass at 15 degrees C, are you sure you were not suppose to record the volume of the saline solution you prepared in lab?

Answer 12

On our lab sheet, all it really says is: Mass of empty 10 mL graduated cylinder ____ Mass of mixture + graduated cylinder ______ Mass of salt_____ etc. The only part I haven't filled in is the Volume of mixture ______ It's the water he gave us with salt in it, we need to find the volume of that.

Answer 13

did he provide you with a density

Answer 14

also this isn't a mixture :S it is a solution

Answer 15

sorry that just really bugs me ha

Answer 16

oh you did this in a graduated cylinder

Answer 17

you should have recorded the final volume after preparing the solution

Answer 18

Yep. a 10 mL graduated cylinder.

Answer 19

so the final volume of the salt solution was 10mL?

Answer 20

Yes.

Answer 21

If you just got a graduated cylinder with 10mL of saline solution in it the volume would be 10mL

Answer 22

which makes my initial calculation incorrect

Answer 23

you would not have 10mL of water you would have something less than 10mL due to displacement caused by the NaCl

Answer 24

Are you serious? It was that simple? :/ I was thinking it would be something totally different.

Answer 25

so you want the percent water easy enough

(Mass of Solute(g)/Volume of Solution(mL)) *100 = %(m/v) concentration of salt in solution
(1g/10mL)*100 = 10%(m/v) salt
thus 10% - 100% = 90%(m/v) water in the solution

Answer 26

Just to clarify

A solution consists of a solute and a solvent, the solute is evenly dispersed in the solvent. If you took 1mL of your 10mL solution you would still have the same concentration of salt in the 1mL

A mixture is an unevenly distribution of chemicals, thus if you removed 1mL of a mixture from a larger mixture you would not have the same concentration in the 1mL mixture as you would in the original mixture

Answer 27

sorry I'm currently experiencing a migraine so my mental faculties are not what I wish they were.

For the future can you please include all the information in your question for the sake of saving the person helping yous time :) not to be a jerk about it.

Answer 28

lol at my subtraction error

Answer 29

In almost all cases like this, you'll assume that the volumes are not additive. The final solution will be 10mL.

Answer 30

Sorry about that. This is my first time actually using this site, so I didn't know what to expect from this. But you helped lots, thanks.

Answer 31

Xishem you cannot say that when the question is stated as having 10 mL of water and 1.0 grams of salt

Answer 32

In any chemistry courses I've taken, the volume addition of a solute is not considered. I can't imagine that a HS-level course would be any different.

Answer 33

Based on that statement you are forced to assume that there are 1g of salt added to 10mL of water

Answer 34

That's exactly what the question says. What are you suggesting it means?

Answer 35

I'm suggesting it means exactly what it sounds like it means. I took it at face value I'm not going to assume she stated her question incorrectly

Answer 36

For all I know the teacher made her record a final volume and she made a mistake in not recording it

Answer 37

I don't even know why we are discussing this

Answer 38

So 1.0g of a salt is being added to 10mL of water. Considering that the salt is unknown, it's impossible to estimate what the total volume displacement is, and in such a case it should not be considered.

Answer 39

I wouldn't say it is impossible you could make a calibration curve of displacement at a fixed temperature and come up with a pretty good estimate

Answer 40

anyway, this problem has been solved so I have no idea why we are arguing