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Chapter 10: Problem 50

Nickel has a face-centered cubic unit cell. The density of nickel is \(6.84 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate a value for the atomic radius of nickel.

Short Answer

Expert verified
The atomic radius of Nickel is approximately \(1.246 \times 10^{-8}\) cm.

Step by step solution

01

Write down known values and formula relating density, edge length, and atomic radius

Given information: Density of Nickel = 6.84 g/cm³ Unit cell: Face-centered cubic (fcc) The formula for density (ρ) is: \(ρ = \frac{mass}{volume}\) For a face-centered cubic unit cell, the number of atoms in the unit cell is 4. We need the formula relating the edge length (a) of the FCC unit cell with the atomic radius (r) which can be derived from geometry: \(a = 2\sqrt{2}r\) We also need the formula relating edge length, mass, and volume: \(\rho = \frac{4 \times mass\ of\ one\ Ni\ atom}{a^3}\)
02

Calculate the molar mass of Nickel

The molar mass of Nickel (Ni) can be found in the periodic table. Molar mass of Ni = 58.69 g/mol
03

Calculate the mass of one Nickel atom

We will use Avogadro's number (6.022 x 10²³ atoms/mol) to determine the mass of one Nickel atom. \(mass\ of\ one\ Ni\ atom = \frac{molar\ mass\ of\ Ni}{Avogadro's\ number}\) \(mass\ of\ one\ Ni\ atom = \frac{58.69 \mathrm{~g/mol}}{6.022 \times 10^{23}\ \mathrm{atoms/mol}} = 9.75 \times 10^{-23}\ \mathrm{g}\)
04

Calculate the edge length of the unit cell

Now, we use the formula for density, which we already derived earlier: \(\rho = \frac{4 \times mass\ of\ one\ Ni\ atom}{a^3}\) Rearrange the formula to find the edge length (a): \(a = \sqrt[3]{\frac{4 \times mass\ of\ one\ Ni\ atom}{\rho}}\) Plug in the density (6.84 g/cm³) and mass of one Ni atom (9.75 x 10^{-23} g) to solve for edge length: \(a = \sqrt[3]{\frac{4 \times 9.75 \times 10^{-23}\ \mathrm{g}}{6.84\ \mathrm{g/cm^{3}}}} = 3.522 \times 10^{-8}\ \mathrm{cm}\)
05

Calculate the atomic radius of Nickel

Use the relationship between the edge length (a) and the atomic radius (r) for a face-centered cubic unit cell: \(a = 2\sqrt{2}r\) Rearrange the formula to find the atomic radius (r): \(r = \frac{a}{2\sqrt{2}}\) Plug in the calculated edge length (3.522 x 10^{-8} cm) to solve for the atomic radius of Nickel: \(r = \frac{3.522 \times 10^{-8}\ \mathrm{cm}}{2\sqrt{2}} = 1.246 \times 10^{-8}\ \mathrm{cm}\) So, the atomic radius of Nickel is approximately 1.246 x 10^{-8} cm.

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Most popular questions from this chapter

Atoms are assumed to touch in closest packed structures, yet every closest packed unit cell contains a significant amount of empty space. Why?

The molar enthalpy of vaporization of water at \(373 \mathrm{~K}\) and \(1.00\) atm is \(40.7 \mathrm{~kJ} / \mathrm{mol}\). What fraction of this energy is used to change the internal energy of the water, and what fraction is used to do work against the atmosphere? (Hint: Assume that water vapor is an ideal gas.)

In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: \(\mathrm{HBr}, \mathrm{Kr}\), or \(\mathrm{Cl}_{2}\) b. highest freezing point: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}\), or HF c. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{Cl}_{2}, \mathrm{Br}_{2}\), or \(\mathrm{I}_{2}\) d. lowest freezing point: \(\mathrm{N}_{2}, \mathrm{CO}\), or \(\mathrm{CO}_{2}\) e. lowest boiling point: \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) f. highest boiling point: HF, HCl, or HBr CC=O g. lowest vapor pressure at \(25^{\circ} \mathrm{C}: \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CCH}_{3}\), or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

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