The molarity of 1 L orthophosphoric acid $$H_3PO_4$$ having 70% purity by weight (specific gravity 1.54 g cm$$^{-3}$$) is ______ M. (Molar mass of $$H_3PO_4 = 98$$ g mol$$^{-1}$$)

We need to find the molarity of 1 L of orthophosphoric acid $$H_3PO_4$$ with 70% purity by weight and specific gravity $$1.54 \, \text{g/cm}^3$$.

First, we recall that molarity is defined as the number of moles of solute per litre of solution: $$M = \frac{\text{Mass of solute in 1 L solution}}{\text{Molar mass of solute}}$$.

Next, the mass of 1 L of solution follows from its density and volume: $$\text{Mass of 1 L solution} = \text{density} \times \text{volume} = 1.54 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1540 \, \text{g}$$.

Since the acid is 70% by weight, the mass of pure $$H_3PO_4$$ in that litre is $$\text{Mass of } H_3PO_4 = \frac{70}{100} \times 1540 = 1078 \, \text{g}$$.

Dividing by its molar mass gives the number of moles: $$\text{Moles} = \frac{1078}{98} = 11.0 \, \text{mol}$$.

Because this amount is present in 1 L of solution, the molarity is $$M = 11.0 \, \text{M}$$.

The correct answer is 11 M.