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A 12.0 M acid solution that contains 75.0% acid by mass has a density of 1.57 g/mL....
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Let the molecular weight of the acid be W.
Hence, 1000 ml 12.0 M acid solution contains 12W grams of pure acid.
Density of the solution = 1.57 g/ml
So, mass of 1000 ml of the solution = (1.57 * 1000) g = 1570 g
Again, mass of the acid in this solution = `75/100*1570` g = 1177.5 g
12W = 1177.5 g
`rArr` W = `1177.5/12` g= 98.125 g
Therefore, the identity of the acid is option (D) `H_3PO_4` (M = 98.0).
Posted by llltkl on August 7, 2013 at 4:33 AM (Answer #1)
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