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

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. 
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?

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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