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?

Answer 1

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.