) 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...
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![### 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:
\[ \text{Mass of one unit cell} = \frac{\text{Molar Mass}}{\text{Avogadro's Number}} \]
\[ \text{Mass of one Po atom} = \frac{208.98 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}} \]
\[ \text{Mass of one Po atom} \approx 3.47 \times 10^{-22} \, \text{g} \]
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