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Chapter 1: Problem 7

From the definition of the mole, show that the molar mass of any element in \(\mathrm{kg}\) is \(10^{-3}\) times its atomic mass in \(\mathrm{u}\), i.c. that the mass of 1 mol in \(\mathrm{g}\) is equal to its atomic mass in \(u\). The A vogadro constant \(N_{A}\) is the number of elementary units of any substance in 1 mol. Show that if \(m_{y}\) is the mass of \(1 \mathrm{u}\) in \(\mathrm{kg}\), then \(N_{A}=1 / 10^{3} m_{v}\) (For convenience, we denote \(10^{3} N_{A}\) in this book by \(N_{A}^{\prime}\) ) Show that a substance with relative atomic or molecular mass \(M\) and density \(\rho\) has a number of atoms or molecules per unit volume equal to \(N_{A} \rho / M\).

Short Answer

Expert verified
Question: Show that a substance with a relative atomic or molecular mass (M) and density (ρ) has a number of atoms or molecules per unit volume equal to (NAρ / M). Answer: To show this, we can follow these steps: 1. Convert the atomic or molecular mass to kg by multiplying by \(10^{-3}\mathrm{kg/g}\). 2. Find the relationship between molar mass and atomic mass (1 mol mass in kg is \(10^{-3}\) times its atomic mass in u). 3. Write an expression for Avogadro's number in terms of the mass of 1 u in kg (\(N_A= \frac{1}{10^3m_y}\)). 4. Derive the formula for the number of atoms or molecules per unit volume using the relationship obtained in step 3: \( \frac{N_A\rho}{M} = \frac{n}{v}\). Thus, a substance with a relative atomic or molecular mass (M) and density (ρ) has a number of atoms or molecules per unit volume equal to (NAρ / M).

Step by step solution

01

Definition and Conversion of Units

The definition of the mole states that one mole of a substance has a mass equal to its atomic or molecular mass in grams. To convert this into kilograms, we can multiply the mass in grams by \(10^{-3}\mathrm{kg/g}\).
02

Molar Mass and Atomic Mass Relationship

If the atomic mass of an element is given in atomic mass units (u), then the mass of 1 mol of the element in grams is equal to its atomic mass in u. Therefore, the mass of 1 mol in kg is \(10^{-3}\) times its atomic mass in u.
03

Finding the Avogadro Constant

From the given information, the mass of 1 u in kg is denoted by \(m_{y}\). Therefore, the mass of Avogadro's number (\(N_A\)) of u in kg is: $$ N_A m_y = 10^{-3}$$ We can solve for Avogadro's number by dividing both sides by \(m_y\): $$ N_A= \frac{1}{10^3m_y}$$
04

Number of Atoms or Molecules per Unit Volume

Now, we want to find the number of atoms or molecules per unit volume for a substance with relative atomic or molecular mass \(M\) and density \(\rho\). The mass of 1 mol of the substance is \(M\) g, and in kg, it will be \(10^{-3}M\) kg. Since there are \(N_A\) atoms or molecules in this mass, the total mass of these atoms or molecules is given by: $$ N_A (10^{-3}M) = \rho v$$ Where \(v\) is the volume occupied by this substance. Now, dividing both sides by the volume \(v\) and the molar mass \(10^{-3}M\), we get: $$ \frac{N_A\rho}{M} = \frac{n}{v}$$ Where \(\frac{n}{v}\) represents the number of atoms or molecules per unit volume. Thus, we have shown that a substance with a relative atomic or molecular mass \(M\) and density \(\rho\) has a number of atoms or molecules per unit volume equal to \(N_{A}\rho / M\).

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Key Concepts

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

Mole
The mole is one of the fundamental concepts in chemistry and is critical for understanding quantitative relationships in chemical reactions. In essence, a mole is a unit of measurement used to express the amount of a substance. It is defined as the number of atoms in exactly 12 grams of pure carbon-12, which is approximately equal to Avogadro's constant, \(6.022 \times 10^{23}\) entities.

Avogadro's constant, \(N_A\), is a dimensionless number representing the number of particles, such as atoms or molecules, in one mole of a substance. Therefore, if you have one mole of water, it contains Avogadro's constant number of water molecules. This makes calculations involving chemical reactions easier, as we often measure substances in terms of their number of moles.
Molar Mass
Molar mass connects the mass of a substance to its mole quantity. It is the mass of one mole of a given substance and the units for molar mass are grams per mole (\(g/mol\)). The molar mass of an element is numerically equivalent to its atomic mass and is expressed in \(g/mol\).

For a compound, the molar mass is the sum of the molar masses of its constituent elements, each multiplied by the number of atoms of that element in the molecular formula. This quantifiable relationship enables chemists to weigh out amounts of a substance on a scale and relate that mass to the number of moles and, in extension, to the number of particles in that substance.
Atomic Mass
Atomic mass is a measure of the mass of a single atom, usually expressed in atomic mass units (u or amu). It roughly corresponds to the sum of the number of protons and neutrons in the atom's nucleus. In this context, carbon-12 is the reference standard, with an atomic mass exactly of 12 u.

Even though the atomic mass is a very small number and difficult to measure in everyday units, it is crucial for scientific calculations. By relating atomic mass units to grams through the mole concept, we can work with measurable quantities in the laboratory.
Molecular Mass
The molecular mass is akin to atomic mass but pertains to molecules. It's the sum of the atomic masses of all atoms in a molecule and is also measured in atomic mass units (u).

For instance, the molecular mass of a water molecule (H2O) is roughly 18 u (16 from oxygen and 1 from each hydrogen atom). By knowing the molecular mass, we can calculate the molar mass, which is the mass of one mole of these molecules in grams, permitting us to convert between moles and grams for that particular compound.
Density
Density is a property that describes how much mass of a substance is contained in a given volume (\(\rho = \frac{m}{V}\), with \(\rho\) being density, \(m\) mass, and \(V\) volume).

It plays a crucial role in the conversion between mass and volume in chemistry. For a substance with a known density, you can determine how much space a certain mass of that substance will occupy, or conversely, the mass associated with a certain volume.
Atoms per Unit Volume
When you know the density of a substance and its molar mass, you can determine the number of atoms or molecules per unit volume. This is a way of expressing concentration that is particularly useful for describing solid materials where 'per liter' or 'per mole' might not be as intuitive.

The formula \( \frac{N_A \rho}{M} \) provides the number of atoms or molecules per unit volume, with \( \frac{n}{v} \) representing the number of atoms or molecules per unit volume. For students working through textbook exercises, understanding this relationship can be critical for solving problems related to the concentration of solutions or the properties of pure substances.

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Most popular questions from this chapter

The charge on a conductor at time \(t\) is given by \(Q=Q_{0}\left(1-e^{-i / T}\right)\) where \(Q_{0}\) and \(T\) are constants. Find the current \(I\) at time \(t\) and sketch the variations of \(Q\) and \(I\) with \(t\). If the charge were given instead by \(Q=Q_{0} e^{-U T}\) how would the current differ from that in the first case?

The current of electrons in a certain cathode-ray tube forms a thin cylindrical beam of constant radius \(a\) in which the charges move parallel to the axis. The current density in a cross-section is found not to be constant but to vary with the distance \(r\) from the axis of the beam according to \(J=J_{0}(1-r / a)\). What is the total current in the beam, assuming that it can be treated as the flow of a continuous distribution of charge? \(J_{0}\) is clearly the current density on the axis at \(r=0\). What is the mean current density in any cross-section in terms of \(J_{0}\) only?

Assuming that each copper atom contributes one free clectronic charge to the current in a wire, estimate the mean drift velocity of charges when the wire has a diameter of \(1 \mathrm{~mm}\) and carries a current of \(1 \mathrm{~A}\). (Relative atomic mass of copper \(=63.6 ;\) density of copper \(=8.9 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\) )

Stow (1969) gives the total transfer of positive charge from the ionosphere to earth due to currents in fine weather as about \(90 \mathrm{C} \mathrm{km}^{-2}\) per year. What is the approximate fine weather current falling on \(1 \mathrm{~mm}^{2}\) of the earth? How many clectronic charges per second does this represent?

The current flowing into a conductor varics in time according to the equation \(l=I_{0} e^{-\alpha !}\), where \(I_{0}\) and \(\alpha\) are constants. Find the charge \(Q\) which has accumulated on the conductor after a time \(t_{0}\). Sketch the variation of \(I\) and \(Q\) with time.
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