Chapter 10.3, Problem 40PP

(a)

Interpretation Introduction

Interpretation:

The number of moles in 22.6 g of ${\text{AgNO}}_{\text{3}}$ needs to be determined.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can be written as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 40PP

$\text{Moles = 0}\text{.133 moles}$

Explanation of Solution

Mass of ${\text{AgNO}}_{\text{3}}$ = 22.6 g

Molar mass of ${\text{AgNO}}_{\text{3}}$ = 169.9 g/mol

Substitute the values to calculate mass:

$\begin{array}{l}\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Moles =}\frac{\text{22}\text{.6 g}}{\text{169}\text{.9 g/mol }}\\ \text{Moles = 0}\text{.133 moles}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The number of moles in 6.50 g of ${\text{ZnSO}}_{\text{4}}$ needs to be determined.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can be written as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 40PP

$\text{Moles = 0}\text{.040 moles}$

Explanation of Solution

Mass of ${\text{ZnSO}}_{\text{4}}$ = 6.50 g

Molar mass of ${\text{ZnSO}}_{\text{4}}$ = 161.5 g/mol

Substitute the values to calculate mass:

$\begin{array}{l}\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Moles =}\frac{\text{6}\text{.50 g}}{\text{161}\text{.5 g/mol }}\\ \text{Moles = 0}\text{.040 moles}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The number of moles in 35.0 g of $\text{HCl}$ needs to be determined.

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can be written as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

Answer to Problem 40PP

$\text{Moles = 0}\text{.959 moles}$

Explanation of Solution

Mass of $\text{HCl}$ = 35.0 g

Molar mass of $\text{HCl}$ = 36.5 g/mol

Substitute the values to calculate mass:

$\begin{array}{l}\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}\\ \text{Moles =}\frac{\text{35}\text{.0 g}}{\text{36}\text{.5 g/mol }}\\ \text{Moles = 0}\text{.959 moles}\end{array}$