The molarity of \(\mathrm{HNO_3}\) in a sample which has density 1.4 g/mL and mass percentage of 63% is _____.
(Molecular Weight of \(\mathrm{HNO_3}\) = 63)
(Molecular Weight of \(\mathrm{HNO_3}\) = 63)
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Correct Answer: 14
Solution and Explanation
Mass of solution = 1000 x 1.4 = 1400 g.
Mass of HNO3 = 63% of 1400 = 882 g.
Moles of HNO3 = 882 / 63 = 14.
Volume = 1 L, so molarity = 14 M.
Answer: 14
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