Calculate molarity and molality of a 21 percent by mass aqueous solution of HNO3 given that density of solution is 1.05 g/ml

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Solution

Dear Student,

Mass of 1 liter of solution:
1.05 g/cm3 times (1000 cm3 / L) = 1050 g/L
Mass of the two components of the solution,

Mass of HNO3 = 21 % of 1050 g = 220.5 g
Mass of H2O =79 % of 1050 g = 829.5 g
Molar mass of HNO3 = 63 g/mol
Moles of Nitric acid = 220.5 g / 63 g/mol = 3.5 mol
Molality = 3.5 mol / 0.829 kg = 4.22 m
Molarity = 3.5 mol / 1.00 L = 3.5 M

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