Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is \(125 \mathrm{pm}\). (a) What is the length of the side of the unit cell? (b) How many unit cells are there in \(1.00 \mathrm{~cm}^{3}\) of aluminium?

Short Answer

Expert verified
The side length of the unit cell is approximately \(250\sqrt{2} \) pm or \(3.54 \times 10^{-8}\) cm. The number of unit cells in \(1.00 \mathrm{cm}^3\) of aluminium is approximately \(6.022 \times 10^{22}\).

Step by step solution

01

Determine the face diagonal in terms of the metallic radius

In cubic close-packed (ccp) or face-centered cubic (fcc) structure, the face diagonal is 4 times the radius of an atom. Let's denote the metallic radius as 'r'. Therefore, the face diagonal 'fd' can be calculated as: \( fd = 4r \).
02

Relate face diagonal to the side length of the unit cell

In a cube, the face diagonal and the side length 'a' are related by the Pythagorean theorem in a right-angled triangle formed by two sides and the face diagonal. Therefore, \( fd = a\sqrt{2} \). By substituting the expression for face diagonal from Step 1, we can solve for 'a': \( 4r = a\sqrt{2} \) , \( a = \frac{4r}{\sqrt{2}} \).
03

Calculate the side length of the unit cell

The metallic radius of aluminium is given as 125 pm. We substitute this value into the expression for 'a': \( a = \frac{4 \times 125}{\sqrt{2}} = \frac{500}{\sqrt{2}} = 250\sqrt{2} \) pm.
04

Convert the side length to centimeters

1 pm = \(1 \times 10^{-12}\) cm. Thus, we convert the side length from pm to cm: \( a = 250\sqrt{2} \times 10^{-12} \) cm.
05

Find the volume of the unit cell

The volume 'V' of the cube (unit cell) is given by the cube of the side length: \( V = a^3 \). Using the value of 'a' from Step 4, calculate the volume.
06

Relate the volume of the unit cell to the given volume of aluminium

To find the number of unit cells in 1.00 \( \mathrm{cm}^3 \), divide the total volume by the volume of a single unit cell: \( \text{number of unit cells} = \frac{1.00 \, \mathrm{cm}^3}{V} \). Calculate this using the volume obtained in Step 5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metallic Radius
The concept of metallic radius is crucial for understanding the arrangement of atoms in a metal. It refers to half the distance between the nuclei of two adjacent atoms in a crystal lattice. In the case of aluminium, which crystallizes in a cubic close-packed (ccp) structure, its metallic radius is given as 125 pm (picometers). This measurement is significant as it determines the size of the unit cell and the arrangement of atoms within the cell.

For metals with a ccp structure, the metallic radius aligns closely with how atoms are tightly packed in the lattice. Since each atom is surrounded by an array of other atoms in this formation, learning about metallic radius gives insights into the density and compactness of the metal. This basic understanding is a stepping stone for detailed calculations related to the unit cell and how it builds up the entire crystal structure of the metal.
Unit Cell Calculation
The unit cell is the smallest repeating unit that composes the crystal structure of a metal and fully defines its crystallography. When we refer to unit cell calculation, we are looking to determine key parameters like the cell's edge length, volume, and the number of atoms per cell. For aluminium, with its given metallic radius, we can compute the length of the edge of the unit cell by employing a geometric relationship involving the face diagonal.

In ccp or fcc structures, the atoms touch along the face diagonals. By knowing that the face diagonal is four times the metallic radius, we can use the Pythagorean theorem to relate the face diagonal to the side length of the cell, which in turn provides the basis to calculate the volume of the unit cell. These computations offer valuable quantitative details about the metal's structure, influencing its properties and practical applications.
Face Diagonal
The face diagonal is an integral line segment within the geometry of the cubic crystal system. It stretches across the face of the cube, connecting opposite corners. In a ccp structure, it is inherently linked to the metallic radius as it is composed of four atomic radii lined up in a straight path across the center of a face.

To calculate the length of the face diagonal, we invoke the metallic radius and the geometry of a right-angled triangle. By understanding this connection, and knowing that the cube's face diagonal is the hypotenuse of a right triangle whose legs are the cube's edges, we can tease out the length of those edges, which define the parameters of the entire crystal lattice. This characteristic face diagonal is instrumental in solving many problems related to the crystalline structure of metals and helps elucidate the relationship between atomic spacing and the properties of the metal.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and equilibrium is represented as : \(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{l})}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}_{(\mathrm{l})} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}\) (a) Write the concentration ratio (reaction quotient), \(Q_{\mathrm{e}}\), for this reaction. Note that water is not in excess and is not a solvent in this reaction. (b) At \(293 \mathrm{~K}\), if one starts with \(1.00\) mole of acetic acid and \(0.180\) of ethanol, there is \(0.171\) mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (c) Starting with \(0.500\) mole of ethanol and \(1.000\) mole of acetic acid and maintaining it at \(293 \mathrm{~K}, 0.214\) mole of ethyl acetate is found after some time. Has equilibrium been reached?

In which case does the reaction go farthest to completion : \(K=1 ; K=10^{10} ; K=10^{-10}\) and why?

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $$ 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NOBr}_{(\mathrm{g})} $$ When \(0.087\) mole of \(\mathrm{NO}\) and \(0.0437\) mole of \(\mathrm{Br}_{2}\) are mixed is a closed container at constant temperature, \(0.0518\) mole of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

The first order diffraction of \(X\) -rays from a certain set of crystal planes oceurs at an angle of \(11.8^{\circ}\) from the planes. If the planes are \(0.281 \mathrm{~nm}\) apart, what is the wavelength of \(X\) -rays?

\(K_{\mathrm{c}}\) for the reaction; \(A+B \Longrightarrow P+Q\), is \(2.0 \times 10^{-2}\) at \(25^{\circ} \mathrm{C}\) and it is \(2.0 \times 10^{-1}\) at \(50^{\circ} \mathrm{C}\). Predict whether the forward reaction is exothermic or endothermic.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free