science(chem)
posted by ~christina~ .
Commercial hydrazine is sold as a 64% solution. Using that as your starting material, how would you prepare a 5% solution of hydrazine?
well if it is sold as 64% solution would that mean that it has 64g of hydrazine per 100ml of solvent(water)?
and what they want in the end is a 5% solution or 5g hydrazine per 100ml of water?
as for how to do this I'm not sure, mathematically or step wise either.
Thanks
science(chem)  DrBob222, Sunday, May 4, 2008 at 6:27pm
percent weight/weight is, in this case,
64 g hydrazine in 100 g SOLUTION. IF, and only if, the density of that solution is 1.00 can that be 100 mL and then it is 100 mL solution and not 100 mL H2O. Since the density of hydrazine is 1.01, I suspect the density of the solution would be so close to 1 that it wouldn't matter; however, since you aren't give a density I assume you are to work the problem not using mL.
A 64% solution is [64 g N2H4/(64 g N2H4 + 36 g H2O)]*100 = 64%.
Now just change the numbers to what you want.
[64 g N2H4/(64 g N2H4 + y g H2O}]*100 = 5% and solve for y g H2O. Then y gH2O + 64 g = total grams of solution.
So you start with 100 g solution(64 g hydrazine) and dilute with water to 1280 g solution and you have a BUNCH of 5% solution. The problem makes it easier by not specifying how much solution is to be prepared. If you don't want that much just divide by an appropriate number to obtain a more manageable quantity. Check my thinking.
I was wondering why is it mass/mass and how do I know if it is? I read on a site that if it isn't specified it would be weight/volume. And I don't understand this part below
"percent weight/weight is, in this case,
64 g hydrazine in 100 g SOLUTION. IF, and only if, the density of that solution is 1.00 can that be 100 mL and then it is 100 mL solution and not 100 mL H2O. Since the density of hydrazine is 1.01, I suspect the density of the solution would be so close to 1 that it wouldn't matter; however, since you aren't give a density I assume you are to work the problem not using mL"
Thanks Dr.Bob

The information I provided is OK. The problem is that we don't know, at least I don't know, if the 64% is a mass/mass, a mass/volume, or whatever. The site may have told you that a solution is mass/volume if it isn't specified, but the information in most texts and all the tutors with whom I have talked say "if no specifics are listed, then it is assumed we are talking about mass/mass percent." And that means, in this case, that 64% is 64 grams of solute in 100 grams of solution. If it is mass/volume, and that is a legitimate way of expressing percent, then 64% means 64 grams solute in 100 mL of solution. Actually, we should always write 64% m/m (usually, however, written as 64% w/w) if it is mass/mass (weight/weight) or 64% m/v to indicate mass/volume percent. The part about the density is this. Often we run across problems in which something is quoted as, say, 0.01% w/w, so we dissolve 0.01 g of solute in 100 grams of solution. However, if the density of the solution is 1.0 g/mL in this dilute solution (the same as water), that allows us to continue to use 0.01 gram solute per 100 grams solution but that also means that it is 0.01 grams solute in 100 mL
With regard to what do you do, I think hydrazine is a liquid, so you need to determine if this is 64% w/v or 64% v/v which would mean 64 g solute in 100 mL solution (for w/v) or 64 mL solute in 100 mL solution (for v/v). Personally, I don't think it makes any difference since the density of hydrazine is very close to 1 (1.01 I think) so 1 g would occupy about 1 mL. Also, I assume this is for an organic experiment and you don't need to be too exact. I hope I've not confused you more. 
no,I understand, but I've tentatively decided to use grams.
But curiousity makes me ask based on the copied statement from above..you have a tutor, Dr.Bob? o.O;;
"but the information in most texts and all the tutors with whom I have talked say"
Thanks for clarifying that Dr.Bob =) 
I have typed a response twice but the board won't let me post it. I don't know the trouble. I'll try this last time.
When I say my tutors I mean those tutors who post on this board, and other boards. I read almost all of them and I learn something new every day. 
Oh..okay.
Respond to this Question
Similar Questions

chem
How would u prepare each of the following solutions 100 mL of a 245 ppm solution of urea, CH4N2O, in water and How would u prepare each of the following solution 100 mL of an aqueous solution whose K+ concentration is 0.084 M Write … 
science(chem)
Commercial hydrazine is sold as a 64% solution. Using that as your starting material, how would you prepare a 5% solution of hydrazine? 
Chemistry
For the reaction of hydrazine (N2H4) in water, Kb is 3.0 106. H2NNH2(aq) + H2O(l) H2NNH3+(aq) + OH (aq) Calculate the concentrations of all species and the pH of a 2.2 M solution of hydrazine in water. okay so i don't know how to … 
chem
You have a large amount of 7.50M stock solution. You need 1.80L of 3.00M solution for an experiment. How would you prepare the desired solution without wasting any stock solution? 
Chemistry
For the reaction of hydrazine (N2H4) in water, Kb is 3.0 106. H2NNH2(aq) + H2O(l)> H2NNH3+(aq) + OH(aq) Calculate the concentrations of all species and the pH of a 1.6 M solution of hydrazine in water. [OH] = [H2NNH3+] = [H2NNH2] … 
Chemisty
Calculate the \rm pH of a 0.10 M solution of hydrazine, \rm N_2H_4. K_b for hydrazine is 1.3\times 10^{6}. 
Chem
I got part a but part b I'm having trouble with I keep getting 56.2mL. I'm not sure what the next step is really to find part b. A. How would you prepare 250 mL of a 0.150 M solution of CoCl2 from solid CoCl2 and distilled water? 
Science ms. Sue please help!
How would you prepare 100.0 mL of 0.07500 M AgNO3 solution starting with pure solute? 
Simple Chemistry
A.Write a brief procedure outling how would you prepare 250mL of a 0.150M solution of a CoCl2 from solid CoCl2 and distilled water b. Using the solution from part a and distilled water, how would you prepare 100ml of a 0.0600 M solution … 
Chemistry
How many mL of 0.0257 N KIO3 would be needed to reach the endpoint in the oxidation of 34.2 mL of 0.0416N hydrazine in hydrochloric acid solution