) The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? (Note: edge length= 2r)
) The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? (Note: edge length= 2r)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
![### Density Calculation of Polonium (Po)
**Problem Statement:**
The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? *(Note: edge length = 2r)*
Considering the atomic radius of Polonium to be 135 pm (picometers), the edge length \(a\) of the simple cubic unit cell would be:
\[ \text{Edge Length (a)} = 2r \]
Given:
\[ r = 135 \, \text{pm} \]
Therefore:
\[ a = 2 \times 135 \, \text{pm} = 270 \, \text{pm} \]
Before proceeding to calculate the density, it is necessary to convert the edge length into centimeters for consistency in the unit used in density calculations.
1 pm \(= 1 \times 10^{-12}\) meters = \(1 \times 10^{-10}\) cm.
Thus:
\[ a = 270 \times 10^{-12} \, \text{m} = 270 \times 10^{-10} \, \text{cm} \]
\[ a = 2.70 \times 10^{-8} \, \text{cm} \]
### Volume of the Unit Cell:
\[ V_{\text{cell}} = a^3 \]
\[ V_{\text{cell}} = (2.70 \times 10^{-8} \, \text{cm})^3 \]
\[ V_{\text{cell}} = 1.97 \times 10^{-23} \, \text{cm}^3 \]
### Mass of One Unit Cell:
The molar mass of Polonium (Po) is approximately 208.98 g/mol. Since a simple cubic unit cell contains only one atom:
\[ \text{Mass of one unit cell} = \frac{\text{Molar Mass}}{\text{Avogadro's Number}} \]
\[ \text{Mass of one Po atom} = \frac{208.98 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]
\[ \text{Mass of one Po atom} \approx 3.47 \times 10^{-22} \, \text{g} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2609b9a-bccc-4ea0-ab3c-7cfd8cf7907c%2Fefe759ce-2974-43a0-9a87-70e19a57ff29%2Fxxon6zm_processed.png&w=3840&q=75)
Transcribed Image Text:### Density Calculation of Polonium (Po)
**Problem Statement:**
The element Polonium (Po) crystallizes in a simple cubic unit cell. If the radius of the Polonium atom is 135 pm, what is the density of Polonium? *(Note: edge length = 2r)*
Considering the atomic radius of Polonium to be 135 pm (picometers), the edge length \(a\) of the simple cubic unit cell would be:
\[ \text{Edge Length (a)} = 2r \]
Given:
\[ r = 135 \, \text{pm} \]
Therefore:
\[ a = 2 \times 135 \, \text{pm} = 270 \, \text{pm} \]
Before proceeding to calculate the density, it is necessary to convert the edge length into centimeters for consistency in the unit used in density calculations.
1 pm \(= 1 \times 10^{-12}\) meters = \(1 \times 10^{-10}\) cm.
Thus:
\[ a = 270 \times 10^{-12} \, \text{m} = 270 \times 10^{-10} \, \text{cm} \]
\[ a = 2.70 \times 10^{-8} \, \text{cm} \]
### Volume of the Unit Cell:
\[ V_{\text{cell}} = a^3 \]
\[ V_{\text{cell}} = (2.70 \times 10^{-8} \, \text{cm})^3 \]
\[ V_{\text{cell}} = 1.97 \times 10^{-23} \, \text{cm}^3 \]
### Mass of One Unit Cell:
The molar mass of Polonium (Po) is approximately 208.98 g/mol. Since a simple cubic unit cell contains only one atom:
\[ \text{Mass of one unit cell} = \frac{\text{Molar Mass}}{\text{Avogadro's Number}} \]
\[ \text{Mass of one Po atom} = \frac{208.98 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]
\[ \text{Mass of one Po atom} \approx 3.47 \times 10^{-22} \, \text{g} \]
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY