$\begin{array}{1 1}16.23M\\18.23M\\17.23M\\20.23M\end{array} $

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68% nitric acid by mass means that

Mass of nitric acid = 68g

Mass of solution = 100g

Molar mass of $HNO_3=63 gmol^{-1}$

$\therefore$ 68 g $HNO_3=\large\frac{68}{63}$ mole =1.079 mole

Density of solution =$1.504gmL^{-1}$

$\therefore$ Volume of solution =$\large\frac{100}{1.504}$mL

$\Rightarrow$ 66.5mL=0.0665 L

Molarity of the solution =$\large\frac{\text{Moles of the solute}}{\text{Volume of solution}}$

$\Rightarrow \large\frac{1.079}{0.0665}$M =16.23M

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