Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Bronx20 on August 05, 2005, 03:50:10 AM
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Hey guys, nice site you have here.
Im stuck on one question and would like some pointers if possible.
The question is, "If you had 1.5 g of NaCl, and you wanted to make a NaCl solution with a concentration of 20 mM, what volume would you have to make the solution up to (NaCl - 58.5 amu)?"
My main problem is getting my head around '20 mM'.
Am I right in assuming that the NaCl is 58.5 g/mol?
Would I then divide the 20 mM by this?
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OK, so I always look at these types of questions by breaking the infomation down to "first principles".
1. m(NaCl) = 1.5g
2. Mw(NaCl) = 58.5g/mol
3. c(NaCl)req = 20mM = 20x10^-3mol/L
and
n(NaCl) = m/Mw = 1.5g/58.5g/mol =
since c(NaCl) = n(NaCl)/V(NaCl) we can rearrange the equation to find the volume,V.
V(NaCl) = (1.5g/58.5g/mol)/(20x10^-3mol/L) = 1.282L or 1282mL
I hope I got this right and have helped.
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Oldddog is right - I used to look at these problems in a similar way:
Remember, you're starting with a fixed number of moles of NaCl. You're just adding water so the number of moles NaCl in the solution won't change. How do you get the # moles; well, just look at the units: molecular weight is in g/mol and mass is in g - if you divide g/(g/mol), you're left with the unit mol. So, to get the # mol, take mass/molecular weight.
Say the # of mol is X, you know that you start with X and finish with X (See above); once again, look at the units of concentration (mol/L) and volume (L); if you multiply the two together (mol/L)*(L) = mol
So, you can set up the equation,
X = (mol/L)*(L) = mol
In other words, the number of moles you start with is the number of moles you finish with
We know that mol/L is concentration (C) and L is volume (V) thus
X = C*V
You know the #mol X and C = is the final concentration you want so to get the volume from above,
V = (X/C)
When doing calculations I make sure to keep a close eye on units – they’ll tell me if I made a mistake
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Excellent, thanks a lot for the replies.
I have always has problems with these questions as I cant work backwards very well.