Concentrated nitric acid has a molarity of 15.9 M and a density of 1.42 g/mL. Calculate the following concentrations using 1.00 L of HNO3.
a) % by mass
b) molality
c) mole fraction

Answers

Answer 1

Answer: a) percent by mass = [tex]70.5\%[/tex]

b) molality = 38.01mol/kg

c) mole fraction = 0.40

Explanation:

Given molarity = 15.9 M

moles of [tex]HNO_3[/tex] = 15.9 moles in 1.00 L of solution

mass of [tex]HNO_3[/tex] = [tex]moles\times {\text {Molar mass}}=15.9mol\times 63g/mol=1001.7g[/tex]

Density of solution of = 1.42 g/ml

Volume of solution = 1.00 L = 1000 ml

Mass of solution = [tex]Density\times volume= 1.42g/ml\times 1000ml=1420g[/tex]

mass of solvent = mass of solution - mass of solute = (1420-1001.7) g = 418.3 g = 0.4183 kg

moles of solvent = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{418.3g}{18g/mol}=23.2mol[/tex]

percent by mass =[tex]\frac{\text {mass of nitric acid}}{\text {mass of solution}}\times 100=\frac{1001.2}{1420}\times 100=70.5\%[/tex]

molality = [tex]\frac{\text {no of moles of solute}}{\text {mass of solvent in kg}}=\frac{15.9mol}{0.4183kg}=38.01mol/kg[/tex]

mole fraction = [tex]\frac{\text {moles of solute}}{\text {total moles}}=\frac{15.9}{15.9+23.2}=0.40[/tex]


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E

Explanation:

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Answer:

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