chemistry

Calculate the molarity of a phosphoric acid(H3PO3)solution that is 84% by mass phosphoric acid and has a density of 1.87g/mL.?

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asked by Lahna
  1. 1.87 g/mL. mass of 1000 mL =
    1.87g/mL x 1000 mL = 1870 grams.
    How much of that is H3PO4? 84%, so
    1870 x 0.84 = grams H3PO4 = 1571g
    How many moles is that?
    1571/molar mass H3PO4 about 16 moles.
    16 moles/L soln = about 16 M.
    I have rounded some of the numbers above; therefore, you need to go through and do it more accurately.

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    posted by DrBob222
  2. How many grams of water and how many grams of 85 wt% phosphoric acid would be needed to make 100 grams of 37 wt% phosphoric acid solution?

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  3. density H3PO4 = 1.689g/L
    density = mass/volume
    mass = 1.689 x 1000
    = 1689g H3PO4 in 1000ml solution

    1)85 wt%
    1689g x 85/100 = 1435.65g H3PO4
    (in 1000ml solution)
    if you want to prepare 100ml of solution
    1000ml = 1435.65g
    100ml = A
    A = 100/1000 x 1435.65g
    = 143.57g H3PO4
    *143.57g H3PO4 needed in prepare 100ml solution of 85wt%

    2)37 wt%
    1689g x 37/100 = 624.93g H3PO4
    (in 1000ml solution)
    if you want to prepare 100ml of solution
    1000ml = 624.93g
    100ml = A
    A = 100/1000 x 624.93
    = 62.49g H3PO4
    *62.49g H3PO4 is needed in prepare 100ml solution of 37%

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    posted by hana

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