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Chapter 13: Problem 16

6.00 M sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\) has a density of \(1.338 \mathrm{g} / \mathrm{mL} .\) What is the percent by mass of sulfuric acid in this solution?

Short Answer

Expert verified
The percent by mass of sulfuric acid in this solution is 43.9%.

Step by step solution

01

Calculate mass of the solute (sulfuric acid)

The molarity of a solution is defined as the number of moles of solute per liter of solution. Therefore, one liter of this 6.00 M sulfuric acid solution contains \(6.00 \times Molar mass of H_{2}SO_{4}\) grams of sulfuric acid. Molar mass of \(H_{2}SO_{4} = (2\times1.0) + 32.0 + (4\times16.0) = 98\,g/mol\), so there are \(6.00 mol/L \times 98\,g/mol = 588\,g\) of \(H_{2}SO_{4}\) in one liter of solution.
02

Calculate the mass of the solvent (water)

The density of the solution is given as 1.338 g/mL. So, the mass of 1 liter of solution (1 L=1000 mL): In grams, this will be \(1.338 g/mL \times 1000 mL = 1338 g\). We previously calculated that 1 L of solution contains 588 g of the solute. Therefore, the mass of water in one liter of solution will be \( 1338 g - 588 g = 750 g\).
03

Calculate percent by mass of sulfuric acid

The percent by mass is given by the mass of the solute divided by the total mass of the solution, times 100. Therefore, \% by mass = \(\frac{Mass_{solute}}{Mass_{solvent} + Mass_{solute}} \times 100%\). Plugging in the values obtained, we find \% by mass = \(\frac{588\,g}{750\,g + 588\,g} \times 100 = 43.9%\). The sulfuric acid solution has a mass/mass percent concentration of 43.9%.

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