Question

Consider the following solids, The solids $A,B,andC$represent aluminum $(D=2.70\mathrm{g}/\mathrm{mL})$, gold )), and silver $(\mathrm{D}=10.5\mathrm{g}/\mathrm{mL})$. If each has a mass of $10.0\mathrm{g}$, what is the identity of each solid? (2.7)

Short answer

The identity of each solid is determined as solid $A$ is gold, solid $B$ is silver, and solid $C$ is aluminum

Step-by-step solution

Introduction 

The given is the solids representing Aluminum, Gold and Silver

The objective is to identify the solids with its appropriate metal

Explanation

Volume of aluminum $=\frac{\text{mass of aluminum}}{\text{density of aluminum}}=\frac{10.0\mathrm{g}}{2.70\mathrm{g}/\mathrm{mL}}=3.70\mathrm{mL}$

Volume of gold $=\frac{\text{mass of gold}}{\text{density of gold}}=\frac{10.0\mathrm{g}}{19.3\mathrm{g}/\mathrm{mL}}=0.52\mathrm{mL}$

Volume of silver $=\frac{\text{mass of silver}}{\text{density of silver}}=\frac{10.0\mathrm{g}}{10.5\mathrm{g}/\mathrm{mL}}=0.95\mathrm{mL}$

As a result, solid $A$ is the gold standard because it takes up less space than solid $B$ and $C$.

Because solid $B$ has a higher volume than solid A but a smaller volume than solid $C$, it takes up more space than $A$.

Because aluminum has a higher volume than gold and silver, it is solid $C$. As a result, it takes up more space than $A$ and $B$.