Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ReVeR on November 08, 2005, 06:08:20 PM
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Hello.
I was wondering what is the molarity of a saturated sodium chloride solution?
Or a way to find it out jsut from the balacned equation....no data given.
I figured it should be about 5.5mol NaCl/Liters of solution.
But this was very rought calculation done at home .
any way to get the excet number without doing an experiment , or anyone know the number?
Thx
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I don't think you can figure it out from the equation but I might be wrong. From wikipedia I got that solubility of NaCl in water is 35.9 g/100 ml (25 °C) (http://en.wikipedia.org/wiki/Sodium_chloride) You just need to convert it to molarity now, but I'll leave that up to you.
How did you get 5.5M?
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problem...i can;'t convert that into the mol/L cuz
it is measured in mols/l of solution!!!!!
solution meaning nacl is in there....all i got from wiki is the amount of water that was there b4 the 35.9 grams were put in. any ideas on coversion from ml of water in ml of solution?
btw 5.5 was a quick test that i did, got to saturation point and over it, then checked how much salt used up etc...
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You can't do it in a single step. First you find out howmany mL of the salt you have. Do that by dividing the density: 35.9g*ml/2.16g=16.62..ml I got the density from Wikipedia. Then I found howmuch solution I have, so 100mL (water) + 16.62=116.62..mL. Now you have amount of solution and amount of the salt so you can go straight into calculations. It turns out to be 5.27M, so you were quite close. This is at 25 celsius of course. I don't know if you can calculate the solubility at different temperatures. I asked my teacher and he dosn't know eighter.
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a solubility of salt is not aeefected in any large way by Temperature i am almsot 100 % sure there...but jsut in case i did do it at 25 C.
However, can we just add the values of volume? i doubt it....
basicly....from my results..
i used 34.8 g of salt.....i almost got to the saturation point, but my final volume was 106 ml.
I am basing the diffrence of volume on the fact that salt dissociated..
and therfore it is nto a straight addition of volums.
any ideas if i had a mistake there or if u can;t do it that way?
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a solubility of salt is not aeefected in any large way by Temperature i am almsot 100 % sure there
Are you sure about this? Usually salts are more soluble at higher temperatures and less soluble at lower temperatures. This is how some recrystalisation works.
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Are you sure about this? Usually salts are more soluble at higher temperatures and less soluble at lower temperatures. This is how some recrystalisation works.
Sodium chloride solubility is 35.7 g/L at 0 deg C and 39.8 g/L at 100 deg C. IIRC this is relatively small difference, it is larger for most salts.
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Sodium chloride solubility is 35.7 g/L at 0 deg C and 39.8 g/L at 100 deg C. IIRC this is relatively small difference, it is larger for most salts.
Ok This is quite small, that certainly is interesting :D Surely not a general rule though?
btw what is IIRC?
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IIRC == If I Recall Correctly
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a solubility of salt is not aeefected in any large way by Temperature i am almsot 100 % sure there...but jsut in case i did do it at 25 C.
However, can we just add the values of volume? i doubt it....
basicly....from my results..
i used 34.8 g of salt.....i almost got to the saturation point, but my final volume was 106 ml.
I am basing the diffrence of volume on the fact that salt dissociated..
and therfore it is nto a straight addition of volums.
any ideas if i had a mistake there or if u can;t do it that way?
Well, I did it the textbook way so you might be right. Also using your data this results in molarity around 5.9-6.0. Is this a different experiment than the one you used for the 5.5 estimate?
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the solubility point of NaCL (soduim chloride=salt) would be 35.9 grams per 100 mL?