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?

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

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  1. 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.

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    posted by DrBob222
  2. I have to go out but I'll check and repost if I don't understand.

    Thanks Dr.Bob

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  3. An ammonia solution is made by diluting 0.150L of the concentrated commercial reagent until the final volume reaches 1.00L.

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    posted by JED

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