Chapter 10, Problem 138A

Interpretation Introduction

(a)

Interpretation:

The number of moles of ions in 0.0200 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:

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

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 138A

Moles of ions = $\text{1}\text{.2}\times {\text{10}}^{\text{-4}}\text{ mole}$

Explanation of Solution

Mass of ${\text{AgNO}}_{\text{3}}$ = 0.0200 g

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

Calculate moles of ${\text{AgNO}}_{\text{3}}$ :

$\text{0}\text{.0200 g ×}\frac{\text{1 mole}}{\text{169}\text{.9 g}}\text{= 1}\text{.2}\times {\text{10}}^{\text{-4}}\text{ mole}$

Since 1 mole of ${\text{AgNO}}_{\text{3}}$ forms 1 mole of ${\text{Ag}}^{+}$ and 1 mole of ${\text{NO}}_{{}_{\text{3}}}^{-}$ ions.

Interpretation Introduction

(b)

Interpretation:

The number of moles of ions in 0.100 moles of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ 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:

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

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 138A

$\begin{array}{l}{\text{K}}^{\text{+}}\text{= 0}\text{.200moles}\\ {\text{CrO}}_{{}_{\text{4}}}^{\text{2-}}\text{= 0}\text{.100 moles}\end{array}$

Explanation of Solution

Moles of ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}$ = 0.100 moles

${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\stackrel{}{\to }{\text{2 K}}^{\text{+}}{\text{+ CrO}}_{{}_{\text{4}}}^{\text{2-}}$

Thus moles of ions:

$\begin{array}{l}{\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\stackrel{}{\to }{\text{2 K}}^{\text{+}}{\text{+ CrO}}_{{}_{\text{4}}}^{\text{2-}}\\ {\text{K}}^{\text{+}}{\text{=2 × moles of K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{=2×0}\text{.100=0}\text{.200moles}\\ {\text{CrO}}_{{}_{\text{4}}}^{\text{2-}}{\text{=moles of K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{= 0}\text{.100 moles}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The number of moles of ions in 0.500 gof ${\text{Ba(OH)}}_{\text{2}}$ 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:

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

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 138A

$\begin{array}{l}{\text{Ba}}^{\text{+}}\text{= 2}{\text{.92×10}}^{\text{-3}}\text{ moles }\\ {\text{OH}}_{}^{\text{-}}\text{= 5}{\text{.84×10}}^{\text{-3}}\text{ moles}\end{array}$

Explanation of Solution

Mass of ${\text{Ba(OH)}}_{\text{2}}$ = 0.500 g

Molar mass of ${\text{Ba(OH)}}_{\text{2}}$ = 171.3 g/mol

Calculate moles:

$\text{0}{\text{.500 g Ba(OH)}}_{\text{2}}\text{× }\frac{\text{1 mole}}{\text{171}\text{.3 g}}\text{= 2}{\text{.92×10}}^{\text{-3}}\text{ moles}$

${\text{Ba(OH)}}_{\text{2}}\stackrel{}{\to }{\text{Ba}}^{\text{+}}{\text{+ 2 OH}}_{}^{\text{-}}$

Thus moles of ions:

$\begin{array}{l}{\text{Ba(OH)}}_{\text{2}}\stackrel{}{\to }{\text{Ba}}^{\text{+}}{\text{+ 2 OH}}_{}^{\text{-}}\\ {\text{Ba}}^{\text{+}}{\text{=moles of Ba(OH)}}_{\text{2}}\text{= 2}{\text{.92×10}}^{\text{-3}}\text{ moles }\\ {\text{OH}}_{}^{\text{-}}{\text{=2 × moles of Ba(OH)}}_{\text{2}}\text{= 2×2}{\text{.92×10}}^{\text{-3}}\text{ moles=5}{\text{.84×10}}^{\text{-3}}\text{ moles}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The number of moles of ions in $1.00\times {10}^{-9}$ moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ 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:

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

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 138A

$\begin{array}{l}{\text{Na}}^{\text{+}}\text{=2}{\text{.00×10}}^{\text{-9}}\text{moles}\\ {\text{CO}}_{{}_{\text{3}}}^{\text{2-}}\text{=1}{\text{.00×10}}^{\text{-9}}\text{moles}\end{array}$

Explanation of Solution

Moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ = $1.00\times {10}^{-9}$ moles

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{2 Na}}^{\text{+}}{\text{+ CO}}_{{}_{\text{3}}}^{2-}$

Thus moles of ions:

$\begin{array}{l}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\to }{\text{2 Na}}^{\text{+}}{\text{+ CO}}_{{}_{\text{3}}}^{\text{2-}}\\ {\text{Na}}^{\text{+}}{\text{=2 × moles of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{=2 ×1}{\text{.00×10}}^{\text{-9}}\text{ =2}{\text{.00×10}}^{\text{-9}}\text{moles}\\ {\text{CO}}_{{}_{\text{3}}}^{\text{2-}}{\text{=moles of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{= 1}{\text{.00×10}}^{\text{-9}}\text{moles}\end{array}$