Question

How can the volume of the largest recorded gold nugget, weighing 93.3 kg and discovered in 1872 in New South Wales, Australia, be calculated in both cubic centimeters (cm³) and liters (L)? Utilize dimensional analysis with units cancellation, considering the density of gold as 19.32 g/mL and the equivalence of 1 cm³ to 1 mL.

Answer 1

The volume of the largest recorded gold nugget, weighing 93.3 kg, is 4825.155 cm³ or approximately 4.825 L, calculated using its mass and the density of gold.

Step-by-step explanation:

To calculate the volume of the largest recorded gold nugget, which weighs 93.3 kg, you can use the density of gold and perform dimensional analysis for units cancellation.

First, convert the mass of the gold nugget from kilograms to grams because the density of gold is given in grams per cubic centimeter (g/cm³).

93.3 kg × 1000 g/kg = 93300 g

Next, use the density of gold, which is 19.32 g/mL (or g/cm³), to find the volume. The formula V = m/d gives us the volume V by dividing the mass m by the density d.

V = 93300 g / 19.32 g/cm³ = 4825.155 cm³

Since 1 mL is equivalent to 1 cm³, the volume in liters (L) is:

V = 4825.155 cm³ × 1 mL/cm³ × 1 L/1000 mL = 4.825155 L

The volume of the gold nugget is 4825.155 cm³ or approximately 4.825 L.