## Chemistry: The Central Science (13th Edition)

$71\%\;HNO_3$ by mass
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 $\frac{1008.2\;g\;HNO_3}{1420\;g\;solution}\times100=71\%\;by\;mass$