Chapter U4, Problem SIII3RE

(a)

Interpretation Introduction

Interpretation:The number of lead ions in the 0.002 M of 10 mL solution of lead nitrate needs to be calculated.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  $\begin{array}{l}\text{Moles = }\frac{\text{mass}}{\text{molar mass}}\\ {\text{1 mole = N}}_{\text{a}}\text{ atoms}\\ {\text{N}}_{\text{a}}\text{ = 6}\text{.022 }\times {\text{ 10}}^{\text{23}}\end{array}$

Answer to Problem SIII3RE

  $\text{1}{\text{.20 ×10}}^{\text{19}}\text{ lead ions}$

Explanation of Solution

Molar mass of lead nitrate = 331.2 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 10²³atoms

Calculate number of lead ions:

  $\text{0}\text{.01L×}\frac{\text{0}\text{.002 mole}}{\text{1 L}}\text{ × }\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1 mole}}\text{× }\frac{\text{1 mole of lead ions }}{\text{1 mole of lead nitrate}}\text{ = 1}{\text{.20 ×10}}^{\text{19}}\text{ lead ions}$

(b)

Interpretation Introduction

Interpretation:The mass of lead nitrate in the 0.002 M of 10 mL solution of lead nitrate needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  $\begin{array}{l}\text{Moles = }\frac{\text{mass}}{\text{molar mass}}\\ {\text{1 mole = N}}_{\text{a}}\text{ atoms}\\ {\text{N}}_{\text{a}}\text{ = 6}\text{.022 }\times {\text{ 10}}^{\text{23}}\end{array}$

Answer to Problem SIII3RE

  $\text{6}{\text{.62 ×10}}^{-3}\text{ g}$

Explanation of Solution

Molar mass of lead nitrate = 331.2 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 10²³atoms

Calculate mass of lead nitrate:

  $\text{0}\text{.01L×}\frac{\text{0}\text{.002 mole}}{\text{1 L}}\text{ × }\frac{\text{331}\text{.2 g}}{\text{1 mole}}\text{ = 6}{\text{.62 ×10}}^{-3}\text{ g}$

(c)

Interpretation Introduction

Interpretation:The mass of lead ion in the 0.002 M of 10 mL solution of lead nitrate needs to be determined.

Concept Introduction: The molecular mass of compound is the sum of atomic mass of all the atoms present in the given chemical compound. Mole concept is used to calculate the moles, mass, number of atoms and moles of a compound. The relation between these values can be shown as:

  $\begin{array}{l}\text{Moles = }\frac{\text{mass}}{\text{molar mass}}\\ {\text{1 mole = N}}_{\text{a}}\text{ atoms}\\ {\text{N}}_{\text{a}}\text{ = 6}\text{.022 }\times {\text{ 10}}^{\text{23}}\end{array}$

Answer to Problem SIII3RE

  $\text{ 4}{\text{.14 ×10}}^{-3}\text{ g}$ lead ions.

Explanation of Solution

Molar mass of lead = 207 g/mol

Concentration = 0.002 M

Volume = 10 mL = 0.01 L

1 mole = 6.022 x 10²³atoms

Calculate mass of lead nitrate:

  $\text{0}\text{.01L×}\frac{\text{0}\text{.002 mole}}{\text{1 L}}{\text{ = 2 ×10}}^{-5}\text{ moles of lead nitrates}$

1 mole of lead ion= 1 mole of lead nitrate

Calculate mass of lead ion:

  ${\text{2 ×10}}^{-5}\text{ moles of lead }ions\text{×}\frac{\text{207 g}}{\text{1 mole}}\text{ = 4}{\text{.14 ×10}}^{-3}\text{ g}$