Question

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

Answer 1

Molar mass of $\text{H}_3\text{PO}_4 = 98 \, \text{g/mol}$

Mass of solution = $1 \times 1000 \times 1.54 = 1540 \, \text{g}$

Mass of $\text{H}_3\text{PO}_4 = 0.7 \times 1540 = 1078 \, \text{g}$

Moles of $\text{H}_3\text{PO}_4 = \frac{1078}{98} = 11 \, \text{moles}$

Thus, molarity = $11 \, \text{M}$.