Dilution with water: from 80% acid to 50% acid (assume 100 cc initial solution): How much water must be added to an 80% solution of sulphuric acid to obtain a 50% solution?

Introduction / Context:
This is a classic dilution (percentage) problem. Because the original question omits the initial volume, we adopt the standard recovery assumption of a 100 cc starting solution for clarity and comparability. The key idea is that the amount of pure acid remains constant while water is added, so the new concentration depends only on the changed total volume.

Given Data / Assumptions:

• Initial volume assumed = 100 cc (Recovery-First standardization).
• Initial concentration = 80% acid.
• Target concentration = 50% acid.
• Water added = w cc (unknown).

Concept / Approach:
Amount of pure acid is conserved during dilution. Set up the conservation equation: initial acid amount = final acid amount. Then solve for the volume of water to be added. Express the final result in cubic centimeters (cc).

Step-by-Step Solution:

Verification / Alternative check:
Final mixture volume = 160 cc with 80 cc acid ⇒ 80/160 = 50%, as required.

Why Other Options Are Wrong:
80, 75, 70, and 50 cc would produce final concentrations of 44.4%, 51.4%, 51.6%, and 61.5% respectively (on a 100 cc starting base), not the desired 50%.

Common Pitfalls:
Treating “percent” as additive rather than conserving the mass of acid, or forgetting to specify/assume an initial volume when the problem statement omits it.

Final Answer:
60 cc