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Math Help - need help fast!!!

  1. #1
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    need help fast!!!

    i hav gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?
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  2. #2
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    Quote Originally Posted by reeree1987 View Post
    i have gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?
    I am pretty sure it's 40 ml of pure hydrochloric acid. If you take the average of 100% and 50% (assuming equal weights of 40 ml each), you will get a 75% acid solution.
    Last edited by mr fantastic; January 22nd 2009 at 02:12 AM. Reason: No edit - just flagging the reply as having been moved from a duplicate post
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  3. #3
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    i hav gotten different ansers for tis poblem
    what "answers" did you get, and how did you get them?


    here is the set-up I would use ...

    let x = amount of pure HCl in mL

    x(100%) + 40(50%) = (x+40)(75%)

    solve for x.
    Last edited by mr fantastic; January 22nd 2009 at 02:15 AM. Reason: No edit - just flagging reply as having been moved from a duplicate post.
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  4. #4
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    Quote Originally Posted by reeree1987 View Post
    i have gotten different ansers for tis poblem "Mow much pure hydrochloric acid must be added to 40 mililiters of a 50% solution to obtain a 75% acid solution?" can anyone clarify this problem for me?

    let:
    x = the amount of pure hydrochloric acid (100%=1)
    (40+x) =total amount

    x(1)+ 40(0.5) = 40+x(0.75)
    x(100%) amount of pure acid + 40ml(50%) solution = 40 + x(75%)
    x+ 20 = 30 + 0.75x
    x - 0.75x = 30 -20
    0.25x = 10
    x= 40 ml


    let's check it
    40(1) + 40(.5) = 40+40(.75)
    40+20= 80(.75)
    60=60
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