0.5 g of fuming (oleum) is diluted with water. The resulting solution requires 26.7 mL of 0.4 N for — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Milli-equivalents of $NaOH$ used = $26.7 \times 0.4 = 10.68$ meq = 0.01068 equivalents. Let the mass of $SO_3$ be $x$ g, so $H_2SO_4$ is $(0.5 - x)$ g. Equivalents of $SO_3$ (n=2, MW=80, E=40) + Equivalents of $H_2SO_4$ (n=2, MW=98, E=49) = 0.01068. $x/40 + (0.5 - x)/49 = 0.01068$. Solving gives $9x = 0.9328 \Rightarrow x \approx 0.1036$ g. Percentage = $(0.1036 / 0.5) \times 100 \approx 20.72\%$.