What is the molar concentration of #68%# #"nitric acid"#, for which #rho_"acid"=1.41*g*mL^-1#?

1 Answer
Aug 28, 2017

#"Molarity"~=15*mol*L^-1#

Explanation:

By definition, #"Molarity"="Moles of solute (mol)"/"Volume of solution (L)"#, and thus it has the units #mol*L^-1#. So we need to address this quotient from the given data......

We assume a #1*mL# volume of solution, the which has a MASS of #1.41*g#, of which 69% of that mass is nitric acid......And so....

#"Molarity"=((1.41*gxx69%)/(63.01*g*mol^-1))/(1.00xx10^-3*L)=15.4*mol*L^-1#

Are you with me......please note the units of the calculation. We wanted an answer with units of #mol*L^-1#, and the quotient gave us such units - and this is an excellent check on our calculations.