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

Step-by-step explanation:

Given molarity = 15.9 M

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

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

Density of solution of = 1.42 g/ml

Volume of solution = 1.00 L = 1000 ml

Mass of solution =
`Density* volume= 1.42g/ml* 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}}=(418.3g)/(18g/mol)=23.2mol`

percent by mass =
`\frac{\text {mass of nitric acid}}{\text {mass of solution}}* 100=(1001.2)/(1420)* 100=70.5\%`

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

mole fraction =
`\frac{\text {moles of solute}}{\text {total moles}}=(15.9)/(15.9+23.2)=0.40`