Question

A solution of \( \text{H}_2\text{SO}_4 \) is 31.4% \( \text{H}_2\text{SO}_4 \) by mass and has a density of \( 1.25 \, \text{g/mL} \). The molarity of the \( \text{H}_2\text{SO}_4 \) solution is _____ M (nearest integer).
[Given molar mass of \( \text{H}_2\text{SO}_4 = 98 \, \text{g/mol} \)]

Answer 1

Correct Answer: 4

To find the molarity (M), use the formula:

\( M = \frac{n_{\text{solute}}}{V} \times 1000 \)

Given that the solution is 31.4% \( H_2SO_4 \), with a density of 1.25 g/mL:

Step 1: Calculate the mass of \( H_2SO_4 \) in 100 g of solution:

Mass of \( H_2SO_4 = 31.4 \, \text{g} \)

Step 2: Convert this to moles:

\( \frac{31.4}{98} = 0.32 \, \text{mol} \)

Step 3: Find the volume of the solution:

\( \text{Volume} = \frac{\text{Mass of solution}}{\text{Density}} = \frac{100}{1.25} = 80 \, \text{mL} \)

Step 4: Calculate molarity:

\( M = \frac{0.32 \, \text{mol}}{80 \, \text{mL}} \times 1000 = 4.005 \approx 4 \, \text{M} \)

So, the correct answer is: 4M

Answer 2

Step 1: Assume 1 L of solution.

Density \( = 1.25\,\text{g/mL} = 1.25\,\text{g/cm}^3 \) \[ \text{Mass of 1 L solution} = 1000\,\text{mL} \times 1.25\,\text{g/mL} = 1250\,\text{g} \]

Step 2: Calculate mass of \( \text{H}_2\text{SO}_4 \).

The solution is 31.4% \( \text{H}_2\text{SO}_4 \) by mass. \[ \text{Mass of } \text{H}_2\text{SO}_4 = \frac{31.4}{100} \times 1250 = 392.5\,\text{g} \]

Step 3: Calculate moles of \( \text{H}_2\text{SO}_4 \).

\[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{392.5}{98} = 4.005 \approx 4\,\text{mol} \]

Step 4: Molarity of the solution.

Molarity \( M = \frac{\text{moles of solute}}{\text{volume of solution in litres}} \) \[ M = \frac{4.0}{1.0} = 4.0\,\text{M} \]

Final Answer:

\[ \boxed{4\,\text{M}} \]