Chapter 13, Problem 13.105PAE

Interpretation Introduction

(a)

Interpretation:

Plated mass of silver should be determined.

Concept introduction:

• Current and time relate the total amount of charge.
• The total amount of charge relates moles of electrons as well.

Answer to Problem 13.105PAE

Solution:

$\text{0}\text{.30}\text{g}$.

Explanation of Solution

Use the given data to calculate the following:

1^st Step: Calculate the total amount of charge applied

$\begin{array}{l}\text{q}\text{=}\text{I × }\text{t}\\ \text{q}\text{=}\text{charge, I = current and t = time}\\ \text{q = 1}\text{.5}\text{A × 3 min ×}\frac{\text{60}\text{s}}{\text{1}\text{min}}\text{=}\text{270}\text{C}\end{array}$

2^nd Step: Calculate moles of electrons

$\begin{array}{l}{\text{1 mol e}}^{\text{-}}\text{=}\text{96500}\text{C}\\ \text{x}\text{mol}{\text{e}}^{\text{-}}\text{=}\text{270}\text{C}\\ \text{x}\text{=}\frac{\text{270}\text{C}}{\text{96500}\text{C/mol}}\text{=}\text{0}\text{.0028}\text{mol}\text{of}{\text{e}}^{\text{-}}\end{array}$

3^rd Step: Calculate total moles of Silver

Since the ratio is 1:1

$\begin{array}{l}{\text{Ag}}^{\text{+}}\text{(aq)}\text{+}{\text{e}}^{\text{-}}\to \text{Ag(s)}\\ \text{mol}\text{of}{\text{e}}^{\text{-}}\to \text{0}\text{.0028}\text{mol}\text{Ag}\end{array}$

4^th Step: Calculate the mass of silver

$\begin{array}{l}\text{mass}\text{=}\text{mol}\text{×}\text{Molar}\text{Mass}\\ \text{mass}\text{=}\text{0}\text{.0028}\text{mol}\text{×}\text{107}\text{.87 }\text{g/mol}\\ \text{mass}\text{=}\text{0}\text{.30 }\text{g}\text{of}\text{Ag}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The number of plated silver atoms should be determined.

Concept introduction:

Use Avogrado’s Number to relate moles and atoms.

Answer to Problem 13.105PAE

Solution:

$\text{1}\text{.69×}{\text{10}}^{\text{21}}\text{atoms}$.

Explanation of Solution

Avogrado’s number relates the total amount of moles and atoms.

Then, use the definition of Avogrado’s number:

$\begin{array}{l}\text{1}\text{mol}\text{=}\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}\\ \text{y}\text{=}\text{mol}\\ \text{y=}\text{0}\text{.0028}\text{mol}\text{×}\frac{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}}{\text{mol}}\\ \text{y}\text{=}\text{1}\text{.69}\text{×}{\text{10}}^{\text{21}}\text{atoms}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The thickness of the plate should be calculated.

Concept introduction:

Use geometric figures, density and masses to relate each dimension.

Answer to Problem 13.105PAE

Solution:

$\text{1}\text{.94}{\text{×10}}^{\text{6}}\text{atoms}$.

Explanation of Solution

This consist of the following steps:

1^st Step: Calculate total Area of the cylinder

$\begin{array}{l}{\text{A}}_{\text{total}}\text{=}{\text{A}}_{\text{wall}}{\text{+A}}_{\text{top}}{\text{+A}}_{\text{bottom}}\\ {\text{A}}_{\text{total}}\text{=}\text{b×h}{\text{+πr}}^{\text{2}}\text{+}{\text{πr}}^{\text{2}}\\ {\text{A}}_{\text{total}}\text{=}\text{πD×h}{\text{+2πr}}^{\text{2}}\\ {\text{A}}_{\text{total}}\text{=}\text{π(3}\text{.00 }\text{mm×2)×(2}\text{.00 }\text{mm)+2π}{\text{(3}}^{\text{.00}}\\ {\text{A}}_{\text{total}}\text{=}\text{92}\text{.25 }{\text{mm}}^{\text{3}}\end{array}$

2^nd Step: Calculate the total volume of the plate

$\begin{array}{l}\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Volume}\text{=}\frac{\text{Mass}}{\text{Density}}\text{=}\frac{\text{0}\text{.30 }\text{g}}{\text{10}\text{.49 }{\text{g/cm}}^{\text{3}}}\text{=}\text{0}\text{.02859 }{\text{cm}}^{\text{3}}\end{array}$

3^rd Step: Calculate the total thickness

$\begin{array}{l}\text{V}\text{=}\text{A×h}\\ \text{h}\text{=}\frac{\text{V}}{\text{A}}\text{=}\frac{\text{0}\text{.028 }{\text{cm}}^{\text{3}}}{\text{92}\text{.25 }{\text{mm}}^{\text{2}}}\text{=}\frac{\text{0}\text{.028 }{\text{cm}}^{\text{3}}}{\text{92}\text{.25 }{\text{mm}}^{\text{2}}\text{×(}\frac{\text{1 }\text{cm}}{\text{10 }\text{mm}}{\text{)}}^{\text{2}}}\text{=}\text{0}\text{.031 }\text{cm}\end{array}$

4^th Step: Calculate the total atoms present

$\begin{array}{l}\text{1 }\text{atm}\text{=}\text{160 }\text{pm}\text{=}\text{1}\text{.6×}{\text{10}}^{\text{-8}}\text{cm}\\ \text{x }\text{atom}\text{=}\text{0}\text{.031 }\text{cm}\\ \text{x}\text{=}\text{0}\text{.031 }\text{cm×}\frac{\text{atom}}{\text{1}\text{.6}{\text{×10}}^{\text{-8}}\text{cm}}\\ \text{x}\text{=}\text{1937500 }\text{atoms}\\ \text{x=}\text{1}\text{.94}{\text{×10}}^{\text{6}}\text{atoms}\end{array}$