Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 13

What is the composition, in atom percent, of an alloy that consists of \(92.5 \mathrm{wt} \% \mathrm{Ag}\) and \(7.5\) wt \(\%\) \(\mathrm{Cu} ?\)

Short Answer

Expert verified
Based on the given information and solution, the alloy's composition in atom percent is approximately 87.9% Ag (silver) and 12.1% Cu (copper).

Step by step solution

01

Obtain molar masses of elements

Look up the molar masses of silver (Ag) and copper (Cu) in the periodic table. The molar mass of Ag is \(107.87\,\text{g/mol}\), and the molar mass of Cu is \(63.55\,\text{g/mol}\).
02

Convert weight percentages to grams

Assume we have \(100\,\text{g}\) of the alloy. Then, we have \(92.5\,\text{g}\) of Ag and \(7.5\,\text{g}\) of Cu.
03

Determine moles of each element in the alloy

Divide the amounts of Ag and Cu in grams by their respective molar masses to find the moles of each element in the alloy. For Ag: \(\text{moles of Ag} = \dfrac{92.5\,\text{g}}{107.87\,\text{g/mol}} = 0.857\,\text{mol}\) For Cu: \(\text{moles of Cu} = \dfrac{7.5\,\text{g}}{63.55\,\text{g/mol}} = 0.118\,\text{mol}\)
04

Calculate total moles in the alloy

Add the moles of Ag and Cu to find the total moles in the alloy. \(\text{Total moles} = 0.857\,\text{mol} + 0.118\,\text{mol} = 0.975\,\text{mol}\)
05

Calculate atom percentages

Divide the moles of each element by the total moles in the alloy and multiply by 100 to obtain atom percentages. For Ag: \(\text{Atom \%} = \dfrac{0.857\,\text{mol}}{0.975\,\text{mol}} \times 100 = 87.9\,\%\) For Cu: \(\text{Atom \%} = \dfrac{0.118\,\text{mol}}{0.975\,\text{mol}} \times 100 = 12.1\,\%\) The composition of the alloy, in atom percent, is approximately \(87.9\,\%\) Ag and \(12.1\,\%\) Cu.

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.

Start your free trial

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

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

Calculate the energy for vacancy formation in nickel (Ni), given that the equilibrium number of vacancies at \(850^{\circ} \mathrm{C}(1123 \mathrm{~K})\) is \(4.7 \times 10^{22} \mathrm{~m}^{-3}\). The atomic weight and density (at \(850^{\circ} \mathrm{C}\) ) for Ni are, respectively, \(58.69 \mathrm{~g} / \mathrm{mol}\) and \(8.80 \mathrm{~g} / \mathrm{cm}^{3}\).

Determine the approximate density of a Ti-6Al-4V titanium (Ti) alloy that has a composition of \(90 \mathrm{wt} \%\) Ti, \(6 \mathrm{wt} \% \mathrm{Al}\), and \(4 \mathrm{wt} \% \mathrm{~V}\)

Calculate the unit cell edge length for an \(80 \mathrm{wt} \% \mathrm{Ag}-\) \(20 \mathrm{wt} \%\) Pd alloy. All of the palladium is in solid solution, the crystal structure for this alloy is \(\mathrm{FCC}\), and the room-temperature density of \(\mathrm{Pd}\) is \(12.02 \mathrm{~g} / \mathrm{cm}^{3} .\)

The concentration of \(\mathrm{Si}\) in an Fe-Si alloy is \(0.25 \mathrm{wt} \%\). What is the concentration in kilograms of Si per cubic meter of allov?

Calculate the number of vacancies per cubic meter in gold (Au) at \(900^{\circ} \mathrm{C}\). The energy for vacancy formation is \(0.98 \mathrm{eV} /\) atom. Furthermore, the density and atomic weight for Au are \(18.63 \mathrm{~g} / \mathrm{cm}^{3}\) (at \(900^{\circ} \mathrm{C}\) ) and \(196.9 \mathrm{~g} /\) mol, respectively.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electromagnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Electricity

Read Explanation

Turning Points in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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