Question

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

Answer 1

`"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.