Question

A 50 mL graduated cylinder contains 25.0 mL of water. A 42.5040 g piece of gold is placed in the graduated cylinder and the water level rises to 27.2 mL. What is the density of the piece of gold?

Answer 1

Answer:

19320 kg/m³

Explanation:

density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.

The formula of density is given as,

D = m/v ......................... Equation 1.

Where D = Density of the gold, m = mass of the gold, v = volume of the gold.

Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.

Amount of water displace = 27.2 - 25 = 2.2 mL.

Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³

Substitute into equation 1

D = 0.042504/0.0000022

D = 19320 kg/m³

Hence the density of the piece of gold = 19320 kg/m³