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Yttrium2
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Preparing a w/w% solution
The formula is grams of solute/grams of solvent x 100
Right?
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gdflp
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Yes, for example a 10% aqueous solution of sodium chloride contains 10g of NaCl and 90g of water per 100g of solution. Try posting these sorts of
simple questions in the Short Questions Thread, or Beginnings not Chemistry in General.
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Yttrium2
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[Edited on 7-2-2015 by Yttrium2]
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blogfast25
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Quote: Originally posted by gdflp  | | Yes, for example a 10% aqueous solution of sodium chloride contains 10g of NaCl and 90g of water per 100g of solution. Try posting these sorts of
simple questions in the Short Questions Thread, or Beginnings not Chemistry in General. |
So w% is weight of solute/(weight of solute + weight of solvent) x 100 %
10 g NaCl / (10 g NaCl + 90 g water) x 100 % = 10 w%
Edit: 10 % of course, NOT 90 %. Typo, my bad.
[Edited on 7-2-2015 by blogfast25]
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gdflp
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| Quote: Originally posted by blogfast25 |
So w% is weight of solute/(weight of solute + weight of solvent) x 100 %
10 g NaCl / (10 g NaCl + 90 g water) x 100 % = 90 w% |
Wow, 90% is pretty concentrated for only adding 10g of NaCl.
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Yttrium2
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Hey now
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Yttrium2
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So if one wanted 100mL of 10% NaCl w/w% how would this be done
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Texium
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I think blogfast must have had a typo when he
said 90%. I think he meant 10%. 10g of solute in 90g of solvent is 10w/w%.
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blogfast25
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If I tell you that the density of water is 1 g/mL, can you work it out from there?
[Edited on 7-2-2015 by blogfast25]
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Yttrium2
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No, wouldn't the density of sodium come into play? Since it isn't 1g/ mL
[Edited on 7-2-2015 by Yttrium2]
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Magpie
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This is the way I would do it:
Let X = grams of NaCl to add to 100g of water to make a 10w% solution:
X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g
Mix 11.1g of NaCl with 100mL water. Pour off 100mL. 
[Edited on 7-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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Strictly speaking yes, but in practice we don't do it that way.
Approx. 100 mL of 10 w% NaCl would mean dissolving 10 g of NaCl in 90g of water. 90 g of water is 90 mL of water. This would give you 100 g of
solution. If the density of that solution is close to 1 g/mL, then 100 g of the solution will also be approx. 100 mL of solution. For most intents and
purposes practical that is good enough.
If you did need to make exactly 100.0 mL of that solution you would have to take into account the density of pure NaCl.
[Edited on 7-2-2015 by blogfast25]
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Yttrium2
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[Edited on 7-2-2015 by Yttrium2]
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Yttrium2
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| Quote: Originally posted by Magpie |
This is the way I would do it:
Let X = grams of NaCl to add to 100g of water to make a 10w% solution:
X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g
Mix 11.1g of NaCl with 100mL water. Pour off 100mL.
Could you explain each step
[Edited on 7-2-2015 by Magpie] |
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blogfast25
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Definitely NOT, you're a making a solution of NaCl in water, NOT Na in water.
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Magpie
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Do you understand basic algebra?
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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| Quote: Originally posted by Yttrium2 | | Quote: Originally posted by Magpie |
This is the way I would do it:
Let X = grams of NaCl to add to 100g of water to make a 10w% solution:
X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g
Mix 11.1g of NaCl with 100mL water. Pour off 100mL.
Could you explain each step
[Edited on 7-2-2015 by Magpie] |
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Did you mean 'pour off 10 mL' ?
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blogfast25
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Me? No. I skipped that bit and went straight to Calculus!
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Magpie
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No. He wanted 100 ml. Realizing that my mixed solution will likely be slightly larger than 100 ml, I must "pour off" only 100ml.
Blogfast that comment about algebra was for the OP. not you!
[Edited on 7-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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Magpie
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Blogfast that comment about algebra was for the OP. not you!
We are responding too fast. This is becoming a cluster-fuck.
[Edited on 7-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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blogfast25
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Jokes aside, in some cases preparing mixtures to an exact volume is necessary of course. In the case of mixers with a precise volume for instance.
100.0 mL of 10 w% NaCl, assuming mixing NaCl with water does not result in contraction or expansion:
Take x g of NaCl and y g of water.
w% of NaCl = x / (x + y) 100 = 10 (eq.1)
Total volume is x/2.165 + y/1.000 = 100 => x + 2.165y = 216.5 (eq.2)
eq.1 => x = 0.1x + 0.1y => 0.9x = 0.1y => x= y/9
Insert into eq.2 => y/9 + 2.165y = 216.5 => 2.28y = 216.5 => y = 95.00 g water
x = 95.00/9 = 10.56 g NaCl
[Edited on 7-2-2015 by blogfast25]
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DistractionGrating
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Well, that's a rather large and quite false assumption. http://bit.ly/1vxPnre
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Yttrium2
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I understand basic algebra, though I do use the book for referencing some things that escape my memory.
just not getting the math! darnit!
will this type of math be included in general chemistry 2?
[Edited on 7-2-2015 by Yttrium2]
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blogfast25
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Not sure how you reach that conclusion.
The link gives density of 1.1972 at 26 w% NaCl (solubility limit IIRW).
26/2.165 + 74 = 86 ml. 100 g/86 ml = 1.163 g/ml calculated, based on no contraction/expansion.
With the listed density of 1.1972, that's a discrepancy of 2.85 % with respect to 1.163. Hardly a "rather large and quite false assumption".
Care to explain?
[Edited on 7-2-2015 by blogfast25]
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Magpie
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Yes. It should have been in general chemistry 1.
[Edited on 7-2-2015 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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