Final answer:
To determine the mass of lead required to displace 234.6 liters of water, we calculate it to be 234600 grams or 234.6 kilograms, based on the density of water and the density of lead using Archimedes' principle.
Step-by-step explanation:
To determine the mass of solid lead that would displace exactly 234.6 liters of water, we need to use the concept of density and Archimedes' principle. The density of water is 1 g/cm³, and since 1 liter is equivalent to 1000 cm³, the mass of water displaced by the lead is 234.6 liters × 1 g/cm³ = 234600 grams (or 234.6 kg).
Now, given that the density of lead is approximately 11.34 g/cm³, we can calculate the mass of lead that would displace this volume of water. By Archimedes' principle, the volume of water displaced is equal to the volume of lead. Therefore, we can calculate the volume of lead as 234600 grams (mass of water) / 11.34 g/cm³ (density of lead) = 20689.24 cm³ (volume of lead).
Finally, to find the mass of lead, we use the volume of lead we just calculated and multiply it by the density of lead: 20689.24 cm³ × 11.34 g/cm³ = 234600 grams, which confirms that the mass of lead needed to displace 234.6 liters of water is 234.6 kilograms.