Question
A sample of nitric acid has a density of 1.42g/cm3 and contains 69.5% by weight of the pure acid.What is the molarity of the solution?
Answers
Devron
Things to keep in mind:
Molarity=Moles/Volume (L)
HNO3=63.01g/mol
69.5%=0.695
and
1cm^3=1mL=10^-3L
Convert Density from g/cm^3 to g/L:
1.42g/cm3*(1cm^3/10^-3L)=1420g/L
Calculate the amount of pure acid:
1.420g/L*(0.695)= mass of pure HNO3 in g/L
Since the molar mass of HNO3=63.01g/mol
Then,
mass of pure HNO3 in g/L*(mol/63.01g)= moles/L=Molarity
***Answer contains no more than 3-significant figures.
Molarity=Moles/Volume (L)
HNO3=63.01g/mol
69.5%=0.695
and
1cm^3=1mL=10^-3L
Convert Density from g/cm^3 to g/L:
1.42g/cm3*(1cm^3/10^-3L)=1420g/L
Calculate the amount of pure acid:
1.420g/L*(0.695)= mass of pure HNO3 in g/L
Since the molar mass of HNO3=63.01g/mol
Then,
mass of pure HNO3 in g/L*(mol/63.01g)= moles/L=Molarity
***Answer contains no more than 3-significant figures.