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chem

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calculate the amount of water that must be added to 6.5 grams of urea to produce a 22-2% by mass solution?

  • chem -

    6.50/g soln) = 0.222
    g soln = 6.50/0.222 = 29.3 g soln
    29.3g soln-6.5g urea = 22.8 g H2O .

    check:
    [6.50g urea/(6.5g urea + 22.8)]*100 =
    [6.50g urea/(29.3g soln)]*100 = 22.18 which rounds to 22.2%

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