Chapter 16, Problem 98A

Interpretation Introduction

Interpretation: The amount (in milliliter) is to be calculated when $1.50\text{ M}$ ${\text{HNO}}_{\text{3}}$ contain enough nitric acid to dissolve an old copper penny with a mass of $3.94\text{g}$ .

Concept Introduction:

The molarity of a substance is defined as the number of moles present in the solute (The substance that is added in a solution is called the solute, e.g., salt is the solute in salt water.). Moles are the units used to represent the smaller units like atoms and molecules and the value of mole is given as $6\cdot 02214076\times {10}^{23}$ .

The expression for moles is as follows:

  $\text{moles}=\frac{{\text{ mass of HNO}}_{\text{3}}}{{\text{Molar mass of HNO}}_{\text{3}}}$   ...... (1)

The expression for molarity is as follows:

  $\text{M}=\frac{\text{moles of solute}}{\text{Volume of solution(L)}}$   ...... (2)

Answer to Problem 98A

The amount $41.66\text{mL}$ of ${\text{HNO}}_{\text{3}}$ is required to dissolve an old copper penny with a mass of $3.94\text{g}$ .

Explanation of Solution

Given:

Molar mass of the solute ${\text{HNO}}_{\text{3}}$ is $63.0\text{g/mol}$ .

Mass of ${\text{HNO}}_{\text{3}}$ is $3.94\text{g}$ .

Molarity is $1.50M$ ${\text{HNO}}_{\text{3}}$ .

Substitute, $3.94\text{g}$ for the mass of ${\text{HNO}}_{\text{3}}$ and $63.0\text{g/mol}$ for molar mass of ${\text{HNO}}_{\text{3}}$ in equation (1).

  $\begin{array}{c}\text{moles}=\frac{\text{ 3}\text{.94 g}}{\text{63}\text{.0 g/mol}}\\ =0.0625\text{ mol}\end{array}$

Substitute, $0.0625\text{ mol}$ for ${\text{HNO}}_{\text{3}}$ and $1.50M$ for molarity of solution in equation (2).

  $\begin{array}{c}\text{Volume of solution}=\frac{\text{Moles solute}}{\text{Molarity}}\\ \text{L}=\frac{0.0625}{1\cdot 50}\\ =0.04166\text{L}\end{array}$

Convert $\text{L}$ to $\text{mL}$ .

  $\begin{array}{c}0.04166\text{L}=0.04166\text{L}\times \frac{1000\text{mL}}{1\text{L}}\\ =41.66\text{mL}\end{array}$

Conclusion

The amount $41.66\text{mL}$ of ${\text{HNO}}_{\text{3}}$ is required to dissolve an old copper penny with a mass of $3.94\text{g}$ .