(N/A) Mass percent of nitric acid in the sample $= 69 \%$.
This means $100 \, g$ of nitric acid solution contains $69 \, g$ of nitric acid by mass.
Molar mass of nitric acid $(HNO_3) = 1 + 14 + 3(16) = 63 \, g \, mol^{-1}$.
Number of moles in $69 \, g$ of $HNO_3 = \frac{69 \, g}{63 \, g \, mol^{-1}} \approx 1.095 \, mol$.
Volume of $100 \, g$ of nitric acid solution $= \frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{100 \, g}{1.41 \, g \, mL^{-1}} \approx 70.92 \, mL = 70.92 \times 10^{-3} \, L$.
Concentration of nitric acid $= \frac{\text{Number of moles}}{\text{Volume in Litres}} = \frac{1.095 \, mol}{70.92 \times 10^{-3} \, L} \approx 15.44 \, mol \, L^{-1}$.