## Chemistry: The Molecular Science (5th Edition)

$30.1\%$
Let the volume of the solution be 1 L. Then, molarity= 10.0 M implies that 10.0 moles of NaOH is in 1 L of solution. Mass of solute=$10.0\,mol\times\frac{39.997\,g}{1\,mol}=399.97\,g$ Mass of the solution= Density$\times$Volume $=\frac{1.33\,g}{1\,cm^{3}}\times\frac{1000\,cm^{3}}{1\,L}\times1\,L=1330\,g$ Weight percent=$\frac{\text{mass of solute}}{\text{mass of solution}}\times100\%$ $=\frac{399.97\,g}{1330\,g}\times100\%=30.1\%$