Answer
$71\%\;HNO_3$ by mass
Work Step by Step
First, find mass of $HNO_3$ (molar mass = 63.01 g/mol) in 1 L of solution.
$\frac{16\;mol\;HNO_3}{1\;L\;solution}(\frac{63.01\;g}{1\;mol})=1008.2\;g\;HNO_3\;per\;L\;solution$
Find mass of 1 L of solution.
$\frac{1\;L\;solution}{}(\frac{1.42\;g}{1\;mL})=1420\;g\;solution\; in\; 1\; L$
Mass percent = mass solute/mass solution x 100
$\frac{1008.2\;g\;HNO_3}{1420\;g\;solution}\times100=71\%\;by\;mass$
