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

Aug 28, 2017

$\text{Molarity} \cong 15 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

By definition, $\text{Molarity"="Moles of solute (mol)"/"Volume of solution (L)}$, and thus it has the units $m o l \cdot {L}^{-} 1$. So we need to address this quotient from the given data......

We assume a $1 \cdot m L$ volume of solution, the which has a MASS of $1.41 \cdot g$, 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 $m o l \cdot {L}^{-} 1$, and the quotient gave us such units - and this is an excellent check on our calculations.