A commercially sold conc. HCl is $$35\%$$ HCl by mass. If the density of this commercial acid is $$1.46$$ g/mL, the molarity of this solution is : (Atomic mass : Cl $$= 35.5$$ amu, H $$= 1$$ amu)

We need to find the molarity of a commercially sold concentrated HCl solution that is 35% HCl by mass with a density of 1.46 g/mL.

The formula for molarity in terms of mass percentage and density is given by:

$$\text{Molarity} = \frac{\text{Mass percentage} \times \text{Density} \times 10}{\text{Molar mass of solute}}$$

The molar mass of HCl is calculated as:

$$M_{HCl} = 1 + 35.5 = 36.5$$ g/mol

Substituting the known values into the formula yields:

$$\text{Molarity} = \frac{35 \times 1.46 \times 10}{36.5}$$

$$= \frac{35 \times 14.6}{36.5}$$

$$= \frac{511}{36.5}$$

$$= 14.0$$ M

To verify this result, consider 1 L (1000 mL) of the solution, for which the mass of solution is:

$$1000 \times 1.46 = 1460$$ g.

The mass of HCl present is 35% of this, namely:

$$0.35 \times 1460 = 511$$ g.

The number of moles of HCl is then:

$$\frac{511}{36.5} = 14.0$$ mol,

confirming a molarity of 14.0 M $$\checkmark$$

Therefore, the correct answer is Option B: 14.0 M.