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

1 Answer
anor277
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.

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