1. Take 1 L of solution

   Since molarity is moles per litre, let us consider $1\,\text{L} = 1000\,\text{mL}$ of the solution.

2. Find mass of 1 L solution using density

   Given density = $1.4\,\text{g mL}^{-1}$

   $\text{Mass of solution} = 1.4 \times 1000 = 1400\,\text{g}$

3. Use mass percentage to find mass of $HNO_3$

   Mass percentage of $HNO_3 = 63\%$

   $\text{Mass of } HNO_3 = \frac{63}{100} \times 1400 = 882\,\text{g}$

4. Convert mass of $HNO_3$ to moles

   Molar mass of $HNO_3 = 63\,\text{g mol}^{-1}$

   $\text{Moles of } HNO_3 = \frac{882}{63} = 14\,\text{mol}$

5. Calculate molarity

   Since these 14 moles are present in $1\,\text{L}$ of solution,

   $\text{Molarity} = \frac{14}{1} = 14\,\text{M}$

Final Answer

$\boxed{14}$