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Chapter 2: Problem 3

(a) How many grams are there in one amu of a material? (b) Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?

Short Answer

Expert verified
Answer: There are 1.66 x 10^{-24} grams in one atomic mass unit of a material, and there are approximately 1.33 x 10^{24} atoms in a pound-mole of a substance.

Step by step solution

01

Understand atomic mass unit (amu) and its relationship with grams

One atomic mass unit (amu) is defined as one twelfth the mass of a carbon-12 atom. We can use this definition to convert amu to grams.
02

Conversion of amu to grams

The mass of one carbon-12 atom is 12 amu, and one amu is equal to 1/12 of that mass. We can calculate the mass of one amu in grams by dividing the mass of one carbon-12 atom in grams by 12. The mass of a carbon-12 atom is approximately 1.99 x 10^{-23} grams, so we have: 1 amu = (1.99 x 10^{-23} grams) / 12 1 amu = 1.66 x 10^{-24} grams (a) There are 1.66 x 10^{-24} grams in one amu of a material.
03

Understand the concept of a mole and Avogadro's number

A mole is a quantity used in chemistry to express amounts of a chemical substance and is usually measured in gram-moles. Avogadro's number (6.022 x 10^{23}) is the number of atoms or molecules in one mole of a substance.
04

Understand the concept of a pound-mole

A pound-mole is a unit of amount of substance used in the United States. It is similar to the gram-mole, but instead of mass in grams, the mass is in pounds. There are 453.592 grams in one pound, so in order to convert from gram-moles to pound-moles, we need to divide by this factor.
05

Calculate the number of atoms in a pound-mole of a substance

Now that we understand what a pound-mole is, we can calculate the number of atoms in a pound-mole using Avogadro's number. 1 pound-mole = (6.022 x 10^{23} atoms or molecules) / 453.592 (b) There are approximately 1.33 x 10^{24} atoms in a pound-mole of a substance.

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Most popular questions from this chapter

To what group in the periodic table would an element with atomic number 114 belong?

The net potential energy \(E_{N}\) between two adjacent ions is sometimes represented by the expression $$ E_{N}=-\frac{C}{r}+D \exp \left(-\frac{r}{\rho}\right) $$ in which \(r\) is the interionic separation and \(C\), \(D\), and \(\rho\) are constants whose values depend on the specific material. (a) Derive an expression for the bonding energy \(E_{0}\) in terms of the equilibrium interionic separation \(r_{0}\) and the constants \(D\) and \(\rho\) using the following procedure: 1\. Differentiate \(E_{N}\) with respect to \(r\) and set the resulting expression equal to zero. 2\. Solve for \(C\) in terms of \(D, \rho\), and \(r_{0}\) - 3\. Determine the expression for \(E_{0}\) by substitution for \(C\) in Equation \(2.12\). (b) Derive another expression for \(E_{0}\) in terms of \(r_{0}, C\), and \(\rho\) using a procedure analogous to the one outlined in part (a).

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