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Kk asked in Science & MathematicsChemistry · 6 years ago

How many mLs of 28% by weight of NH2(17.0g/mol) with a density of 0.890 g/ml are needed to prepare 500mL?

Of 0.250 M concentration. I got 8.53 but my answer is wrong. How do you get the correct answer ?

2 Answers

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  • 6 years ago
    Favorite Answer

    mass of NH3 required:

    (0.250 mol/L) (0.5 L) = x / 17.0 g/mol

    x = 2.125 g

    mass of 28% NH3 that contains 2.125 g:

    2.125 is to y as 28 is to 100

    y = 7.589286 g

    volume of 7.589286 g of 28% solution:

    7.589286 g divided by 0.890 g/mL = 8.53 g

    Your answer is correct. So, why was it called wrong? One possibility is that 8.5 is the expected answer (you had too many sig figs for the computer). Another possibility is that the answer key in the computer is wrong. Is there a live person who can access the answer in the computer and confirm/disconfirm this?

    Best wishes with your continued studies.

    Source(s): ChemTeam
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  • 6 years ago

    The original stock solution of NH3:

    28% means that 100 g of the solution will contain 28 g of NH3. Moles NH3 in that volume is: 28g / 17.0 g/mol = 1.65 moles NH3

    The volume of 100 g of solution is: 100 g / 0.89 g/mo = 112 mL or 0.112 L

    Molarity of stock solution = 1.65 mol 0.112 L = 14.7 M

    Now,

    M1V1 = M2V2

    14.7 M ( V1) = 500 mL (0.25M)

    V1 = 8.50 mL

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