A chemical supply company sells a concentrated solution of aqueous H2SO4 (molar mass 98 g mol−1) tha — Solutions Chemistry Question

STEPS:

1. Choose a convenient basis for calculation: Assume a volume of exactly $1.0\text{ L}$ (or $1000\text{ mL}$) of the concentrated $\text{H}_2\text{SO}_4$ solution to make the calculations straightforward.
2. Calculate the total mass of this solution volume: Use the given density of the solution ($1.4\text{ g mL}^{-1}$) to convert the volume of the solution to its total mass:
$$\text{Total mass of solution} = \text{Volume} \times \text{Density} = 1000\text{ mL} \times 1.4\text{ g mL}^{-1} = 1400\text{ g}$$
3. Determine the mass of the $\text{H}_2\text{SO}_4$ solute: Since the solution is $50.\text{ percent }\text{H}_2\text{SO}_4$ by mass, calculate the mass of the actual solute in the solution:
$$\text{Mass of }\text{H}_2\text{SO}_4 = 1400\text{ g} \times 0.50 = 700\text{ g}$$
4. Calculate the moles of $\text{H}_2\text{SO}_4$ solute: Divide the mass of the solute by its molar mass ($98\text{ g mol}^{-1}$):
$$\text{Moles of }\text{H}_2\text{SO}_4 = \frac{700\text{ g}}{98\text{ g mol}^{-1}} \approx 7.14\text{ mol}$$
5. Determine the molarity of the solution: Molarity ($M$) is defined as the moles of solute dissolved per liter of solution. Since our initial basis was $1.0\text{ L}$ of solution:
$$\text{Molarity} = \frac{7.14\text{ mol}}{1.0\text{ L}} \approx \mathbf{7.1\text{ M}}$$
This matches Option D.

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WHY_OTHERS_WRONG:

• A is incorrect: This value ($1.8\text{ M}$) is half of the incorrect calculation in Option B, which occurs if a student mistakenly applies the $50\%$ mass percent factor a second time or makes a calculation error while scaling the moles of solute.
• B is incorrect: This value ($3.6\text{ M}$) is obtained if a student incorrectly multiplies by the density instead of dividing when converting solution mass to volume (e.g., using a $100\text{ g}$ basis: incorrectly calculating solution volume as $100\text{ g} \times 1.4\text{ g mL}^{-1} = 140\text{ mL}$, which yields $\frac{0.51\text{ mol}}{0.14\text{ L}} \approx 3.6\text{ M}$).
• C is incorrect: This value ($5.1\text{ M}$) occurs if a student confuses mass percent ($\text{w/w}$) with mass/volume percent ($\text{w/v}$) and completely neglects the density of the solution (e.g., assuming there are exactly $50\text{ g}$ of solute in $100\text{ mL}$ of solution: $\frac{50\text{ g} / 98\text{ g mol}^{-1}}{0.100\text{ L}} \approx 5.1\text{ M}$).
• E is incorrect: This value ($14\text{ M}$) represents the molarity of a hypothetical $100\%$ pure $\text{H}_2\text{SO}_4$ liquid with a density of $1.4\text{ g mL}^{-1}$ ($\frac{1400\text{ g} / 98\text{ g mol}^{-1}}{1.0\text{ L}} \approx 14.3\text{ M}$), which fails to account for the $50\%$ water dilution in this aqueous solution.