Q. 2.91 - An Introduction to General, Organic, and Biological Chemistry

Question

Consider the following solids, The solids $A, B, and C$ 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)

Step-by-Step Solution

Answer

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

Step 1 : Introduction

The given is the solids representing Aluminum, Gold and Silver

The objective is to identify the solids with its appropriate metal

Step 2 : 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 $B$.