Calculate the volume of nitric acid (65%, density=1.40g/cm3) and how cm3 H2O need to added 500cm3 solution 0.2mol/dm3.Calculate the quantitative participation(Xa=nA/nA+nB) of the initial solution.

Question

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Answer 1

The question, as written, is confusing. Assuming you are trying to make 500 ml of 0.2 M nitric acid, then we can proceed as follows:

moles HNO3 needed = 500 ml x 1 L / 1000 ml x 0.2 mol / L = 0.100 moles HNO3 needed

grams HNO3 needed = 0.100 mols HNO3 x 63.0 g / mol = 6.30 g HNO3

grams of 65% HNO3 needed = 6.30 g / 0.65 = 9.69 g of solution

volume 65% HNO3 needed = 9.69 g x 1 ml / 1.40 g = 6.92 mls of 65% HNO3

volume of H2O needed = 500 ml - 6.92 ml = 493 mls

To find mole fraction of HNO3 (X_HNO3) divide moles HNO3 by moles HNO3 + moles H2O:

moles HNO3 = 0.100 moles

moles H2O = 493 mls H2O x 1 g / ml x 1 mol / 18 g = 27.4 mols

Total moles = 0.100 + 27.4 = 27.5

X_HNO3 = 0.1 / 27.5 = 0.0036