Question

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

Answer 1

Answer: a) percent by mass = $70.5\%$

b) molality = 38.01mol/kg

c) mole fraction = 0.40

Explanation:

Given molarity = 15.9 M

moles of $HNO_3$ = 15.9 moles in 1.00 L of solution

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

Density of solution of = 1.42 g/ml

Volume of solution = 1.00 L = 1000 ml

Mass of solution = $Density\times volume= 1.42g/ml\times 1000ml=1420g$

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

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

percent by mass =$\frac{\text {mass of nitric acid}}{\text {mass of solution}}\times 100=\frac{1001.2}{1420}\times 100=70.5\%$

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

mole fraction = $\frac{\text {moles of solute}}{\text {total moles}}=\frac{15.9}{15.9+23.2}=0.40$