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

To find the molarity of orthophosphoric acid (\( H_3PO_4 \)) given 70% purity by weight and specific gravity of \(1.54 \, \text{g cm}^{-3}\), follow these steps:

1. Determine the density of the acid:
Specific gravity is the ratio of the substance's density to that of water. Thus, density of \( H_3PO_4 \) is:
\( 1.54 \times 1 \, \text{g cm}^{-3} = 1.54 \, \text{g cm}^{-3} = 1540 \, \text{g L}^{-1} \) (since \(1 \, \text{cm}^3 = 1 \, \text{mL}\) and \(1000 \, \text{mL} = 1 \, \text{L}\)).

2. Calculate mass of \( H_3PO_4 \) in 1 L solution:
Given 70% purity by weight, the mass of \( H_3PO_4 \) in the solution is \( 0.7 \times 1540 \, \text{g} = 1078 \, \text{g} \).

3. Convert mass to moles:
The number of moles of \( H_3PO_4 \) is given by \(\frac{1078 \, \text{g}}{98 \, \text{g mol}^{-1}} = 11 \, \text{mol} \).

4. Calculate molarity:
Molarity is moles of solute per liter of solution, which is simply \(11 \, \text{M}\) since we calculated 11 moles in 1 L.

Thus, the molarity of the solution is 11 M.

Finally, we verify this value against the range of 11–11. The computed molarity is exactly 11 M, which fits perfectly within the specified range.

Answer 2

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