An analytical laboratory analyzes a commercial sample of fuming sulphuric acid (oleum) tightly label — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

Step 1: Interpret the Oleum label. A strength of "$109\%$" chemically indicates that $100\text{ g}$ of this oleum requires the addition of exactly $9\text{ g}$ of pure water to convert all the dissolved "free" $SO_3$ into $H_2SO_4$, resulting in a final mass of $109\text{ g}$ of pure $H_2SO_4$. Step 2: Write the hydration reaction. $H_2O(l) + SO_3(g) \rightarrow H_2SO_4(l)$. Step 3: Perform stoichiometric calculation. The balanced equation shows that $18\text{ g}$ of water ($1\text{ mole}$) reacts stoichiometrically with exactly $80\text{ g}$ of $SO_3$ ($1\text{ mole}$). Therefore, $9\text{ g}$ of water will strictly react with $\left(\frac{80}{18}\right) \times 9 = 40\text{ g}$ of $SO_3$. This represents the mass of "free" $SO_3$ initially present.