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

Correct Answer: 11

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 = \(11 \, \text{M}\).