Chapter 8, Problem 17E

Interpretation Introduction

(a)

Interpretation:

The mass in grams for $\text{2}\text{.95}\times {\text{10}}^{\text{23}}$ atoms of mercury, $\text{Hg}$ is to be stated.

Concept introduction:

A mole is the amount of substance that has Avogadro’s number of particles. The Avogadro’s number is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$. Mathematically, the mole is the ratio of the mass and the molecular mass of the substances.

$\begin{array}{rll}\text{Mole}& =& \frac{\text{Mass}}{\text{Molecular mass}}\\ \text{Mass}& =& \text{Mole}\times \text{Molecular mass}\end{array}$

One mole of atom or molecule is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$ particles.

Interpretation Introduction

(b)

Interpretation:

The mass in grams for $1.16\times {10}^{22}$ molecules of nitrogen, ${\text{N}}_{\text{2}}$ is to be stated.

Concept introduction:

A mole is the amount of substance that has Avogadro’s number of particles. The Avogadro’s number is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$. Mathematically, the mole is the ratio of the mass and the molecular mass of the substances.

$\begin{array}{rll}\text{Mole}& =& \frac{\text{Mass}}{\text{Molecular mass}}\\ \text{Mass}& =& \text{Mole}\times \text{Molecular mass}\end{array}$

One mole of atom or molecule is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$ particles.

Interpretation Introduction

(c)

Interpretation:

The mass in grams for $5.05\times {10}^{21}$ formula units of barium chloride, ${\text{BaCl}}_{\text{2}}$ is to be stated.

Concept introduction:

A mole is the amount of substance that has Avogadro’s number of particles. The Avogadro’s number is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$. Mathematically, the mole is the ratio of the mass and the molecular mass of the substances.

$\begin{array}{rll}\text{Mole}& =& \frac{\text{Mass}}{\text{Molecular mass}}\\ \text{Mass}& =& \text{Mole}\times \text{Molecular mass}\end{array}$

One mole of atom or molecule is equal to $\text{6}\text{.02}\times {\text{10}}^{\text{23}}$ particles.