1. What volume of 37% Formaldehyde solution must be added to 1 liter of water to obtain 1g of CH2O per liter?
2. Same question above but with 40% Formaldehyde.
3. Same question but with 10% Formalin.DDTea - 5-2-2011 at 17:22
Is this a homework question? When in doubt, calculate the number of moles! That's always a good place to start. Also, please make some attempt at
answering this; no one is going to just give you an answer and that wouldn't be helpful to you either. It's better if we can see where you're having
trouble and take it from there.
jonco - 5-2-2011 at 17:30
1. I took the molar mass 30.03g and multiplied by 37% and also by 100ml/1000ml for percentage concentration and I got 1.1ml
2. I did the same for 40% and I got 1.2ml
3. I used 10% because formalin is different name for formaldehyde so it was tricky but I got it right with using same formula for other 2. So I should
use 0.3ml for this one.
But I don't know if this is right. DDTea - 5-2-2011 at 17:34
It's not right; do a quick reality check. According to your answer, you would use less volume of a dilute Formaldehyde solution to achieve
the same concentration in 1 L of water. That doesn't make sense.
Try to convert % concentration into molarity (Moles/Liter). Also, find the molarity of a 1g CH2O solution in 1 L of water. From there, you can use
the equation "(M1)(V1) = (M2)(V2)"
Where M1 = molarity of conc. solution, V1 = volume of conc. solution, M2 = molarity of dilute solution, V2 = Volume of dilute solution. You will be
solving V1. entropy51 - 5-2-2011 at 17:47
Solving V1 = 1000 C2 / (C2 - C1) = 1000 x 1 / (370 - 1) = 2.7 mL
You get the same result by taking V2 = 1000 because V1 << 1000.
There is no need to convert to moles because the initial and final concentrations are both in g/L. You need moles for neutralizations (titrations)
but not necessarily for simple dilution problems.
I probably screwed up the algebra.
[Edited on 6-2-2011 by entropy51]jonco - 5-2-2011 at 17:57
Since you wanted 1 g/L, your answer is not correct, assuming that C1 = 37% is gm CHO per mL of solution.
But if C1 = 37% is gm CHO / gm solution, then convert using density C1 = 0.37 x 1.38 = 0.51 gm CHO / mL or 510 gm CHO per liter so that C2 = 510
(.00196) / 1 = 0.999 and your answer is correct.
If the units on % are not stated, more often than not the weight percent is what is being given (gm solute/gm soln).
[Edited on 6-2-2011 by entropy51]jonco - 6-2-2011 at 03:23
Don't we need the density (1.38 g/mL) of 37% Formaldehyde to convert from percentage to molarity, or does 37% always equal 37g CH2O per 100ml?unionised - 6-2-2011 at 07:21
The question is not properly specified
Also the density of formalin isn't that high, it's wrong on Wiki too if anyone cares to change it..jonco - 6-2-2011 at 07:32
Why is the question not properly specified? Is there insufficient information to solve it?jonco - 19-2-2011 at 10:48
Does anyone know how to solve the original problem in the first post?Magpie - 19-2-2011 at 11:09
Does anyone know how to solve the original problem in the first post?
For 1:
(0.37)(sp gr)V = 1
Where sp gr = the specific gravity of the 37% formaldehyde
and V= the volume in mL. Take the V and add it to enough water to bring the total volume to 1 liter.
You will have to look up the sp gr of the formalin.
[Edited on 19-2-2011 by Magpie]jonco - 2-3-2011 at 12:50
sp gr = 1.08 for 40% Formaldehyde
V = 1g/(0.4*1.08) = 2.315mL
Add 2.315mL 40% formaldehyde to 997.685mL of water to get 1g CH2O per litre.
Can someone confirm if this is the correct answer?jonco - 28-5-2011 at 03:28
Bump
Can someone tell me if the answer is correct?Magpie - 28-5-2011 at 16:27
Yes, that's basically correct. In general, you should add the calculated volume to a vessel with a 1L mark. Then add enough water to bring the level
up to the mark. This eliminates any errors caused by volume changes due to mixing.
