Solutions Part 2

Concept:Mole fraction is the number of moles of a component divided by the total number of moles in the solution.Explanation:First, find the mass of pure $\mathrm{H}_2\mathrm{SO}_4$ contributed by each solution.First solution: $100\ \text{g} \times 98\% = 98\ \text{g}$.Second solution: $100\ \text{g} \times 49\% = 49\ \text{g}$.Total mass of $\mathrm{H}_2\mathrm{SO}_4$ = $98 + 49 = 147\ \text{g}$.Total mass of the final solution = $100 + 100 = 200\ \text{g}$.Mass of water in the mixture = $200 - 147 = 53\ \text{g}$.Molar mass of $\mathrm{H}_2\mathrm{SO}_4$ = $2 + 32 + 64 = 98\ \text{g mol}^{-1}$.Moles of $\mathrm{H}_2\mathrm{SO}_4$ = $\frac{147}{98} = 1.5\ \text{mol}$.Molar mass of $\mathrm{H}_2\mathrm{O}$ = $18\ \text{g mol}^{-1}$.Moles of $\mathrm{H}_2\mathrm{O}$ = $\frac{53}{18} \approx 2.944\ \text{mol}$.Mole fraction of $\mathrm{H}_2\mathrm{SO}_4$ = $\frac{1.5}{1.5 + 2.944} = \frac{1.5}{4.444} \approx 0.337$.Answer:$0.337$ (Option D).