Question

a gold block, with a mass of 386 g, will displace 20 ml of water in a graduated cylinder. the diagram below shows what happens when an unknown, irregular solid, also with a mass of 386 g is immersed in water. using the method described in the passage, compare the density of the unknown with the density of gold.

Explanation:

Step1: Calculate the density of gold

The formula for density is \(
ho=\frac{m}{V}\). For gold, \(m = 386\space g\) and \(V = 20\space mL\). So, \(
ho_{gold}=\frac{386}{20}=19.3\space g/mL\)

Step2: Calculate the volume of the unknown solid

The initial volume of water is \(50\space mL\) and the final volume is \(60\space mL\). So the volume of the unknown solid \(V_{unknown}=60 - 50=10\space mL\)

Step3: Calculate the density of the unknown solid

Using the density formula \(
ho=\frac{m}{V}\), with \(m = 386\space g\) and \(V = 10\space mL\). So, \(
ho_{unknown}=\frac{386}{10}=38.6\space g/mL\)

Step4: Compare the densities

Since \(38.6>19.3\)

Answer:

The density of the unknown solid (\(38.6\space g/mL\)) is greater than the density of gold (\(19.3\space g/mL\))