The molarity of $$HNO_3$$ in a sample which has density 1.4 g/mL and mass percentage of 63% is ___________.
(Molecular Weight of $$HNO_3 = 63$$)

We start by recalling that density tells us the mass of the solution per unit volume. The given density is $$1.4\ \text{g mL}^{-1}$$. Because there are $$1000\ \text{mL}$$ in $$1\ \text{L}$$, we convert this to grams per litre:

$$1.4\ \text{g mL}^{-1}\times 1000\ \text{mL L}^{-1}=1400\ \text{g L}^{-1}.$$

This means that one litre of the solution has a total mass of $$1400\ \text{g}$$.

Next, the mass percentage of nitric acid is $$63\%$$. By definition of mass percentage, $$63\%$$ $$\bigl(\text{w/w}\bigr)$$ means that in every $$100\ \text{g}$$ of the solution, the mass of $$HNO_3$$ present is $$63\ \text{g}$$. Therefore, we set up a direct proportion for one litre of the solution:

$$\frac{63\ \text{g HNO}_3}{100\ \text{g solution}}=\frac{x\ \text{g HNO}_3}{1400\ \text{g solution}}.$$

Cross-multiplying, we obtain

$$x=1400\times\frac{63}{100}=1400\times0.63=882\ \text{g HNO}_3.$$

So, one litre of the solution contains $$882\ \text{g}$$ of nitric acid.

To find molarity, we need the number of moles of $$HNO_3$$ present in this one-litre volume. The formula for moles is stated first:

$$\text{Moles}=\frac{\text{Mass}}{\text{Molar Mass}}.$$

The molar mass (molecular weight) of $$HNO_3$$ is given as $$63\ \text{g mol}^{-1}$$. Substituting the values:

$$\text{Moles of }HNO_3=\frac{882\ \text{g}}{63\ \text{g mol}^{-1}}=14\ \text{mol}.$$

Because these $$14$$ moles are present in exactly $$1\ \text{L}$$ of solution, the \b{Molarity} $$M$$ (moles per litre) is simply

$$M=14\ \text{mol L}^{-1}.$$

So, the answer is $$14$$.