Problem 2
Question
A diatomic molecule \(X_{2}\) has a body-centred cubic \((b c c)\) structure with a cell edge of \(300 \mathrm{pm}\). The density of the molecule is \(6.17 \mathrm{~g} \mathrm{~cm}^{-3}\). The number of molecules present in \(200 \mathrm{~g}\) of \(\mathrm{X}_{2}\) is : (Avogadroconstant \(\left.\left(\mathrm{N}_{\mathrm{A}}\right)=6 \times 10^{23} \mathrm{~mol}^{-1}\right)\) (a) \(40 \mathrm{~N}_{\mathrm{A}}\) (b) \(8 \mathrm{~N}_{\mathrm{A}}\) (c) \(4 \mathrm{~N}_{\mathrm{A}}\) (d) \(2 \mathrm{~N}_{\mathrm{A}}\)
Step-by-Step Solution
Verified Answer
The number of molecules is \(4 \text{N}_{\text{A}}\).
1Step 1: Understand the Structure
A body-centred cubic (bcc) structure has one additional atom at the body center of the cube, apart from the eight corner atoms. Thus, for the bcc lattice, the effective number of atoms per unit cell is 2.
2Step 2: Calculate Volume of the Unit Cell
The volume (
V
) of a unit cell can be calculated from the cell edge length (
a
) using the formula
V = a^3
. Here,
a = 300 ext{ pm} = 300 imes 10^{-10} ext{ cm}
. Therefore,
V = (300 imes 10^{-10})^3 ext{ cm}^3 = 27 imes 10^{-24} ext{ cm}^3
.
3Step 3: Calculate Mass of the Unit Cell
The mass of the unit cell can be found using the formula
ext{Mass} = ext{Density} imes ext{Volume}
. Substituting the given values,
ext{Mass} = 6.17 ext{ g/cm}^3 imes 27 imes 10^{-24} ext{ cm}^3 = 1.6659 imes 10^{-22} ext{ g}
.
4Step 4: Calculate Molar Mass
Since the unit cell contains two atoms of
X_2
, we can write
2m = 1.6659 imes 10^{-22} ext{ g}
where
m
is the molar mass of
X_2
. This gives
m = 1.6659 imes 10^{-22} / 2 = 0.83295 imes 10^{-22} ext{ g}
. To convert to
g/mol
, use
M = m imes N_A = 0.83295 imes 10^{-22} imes 6 imes 10^{23}
, which simplifies to
M = 49.977 ext{ g/mol}
.
5Step 5: Calculate Number of Moles in 200g
We calculate the number of moles (
n
) in 200g using the formula
n = ext{mass} / ext{molar mass}
. Hence,
n = 200 / 49.977 ext{ mol} = 4 ext{ mol}
.
6Step 6: Calculate Number of Molecules
The number of molecules is given by
ext{Number of molecules} = n imes N_A
where
n = 4 ext{ mol}
. Therefore, the number of molecules is
4 imes 6 imes 10^{23} = 24 imes 10^{23} ext{ molecules}
, which is equivalent to
4 ext{N}_{ ext{A}}
.
Key Concepts
Diatomic MoleculeUnit Cell VolumeMolar Mass Calculation
Diatomic Molecule
A diatomic molecule consists of two atoms bonded together. In this exercise, the molecule being discussed is represented as \(X_2\). It implies that the compound is made up entirely of two atoms of the element \(X\). Diatomic molecules are quite common, with oxygen \(\text{O}_2\) and hydrogen \(\text{H}_2\) being well-known examples. These molecules exist because the pairing of atoms provides stability through shared electrons in chemical bonds.
Understanding the diatomic nature of a molecule helps in calculating properties such as molar mass, especially when analyzing crystal structures like the body-centred cubic (bcc) structure. The internal arrangement of atoms in a bcc lattice, where each unit cell effectively contains two atoms, aligns well with the diatomic configuration that allows specific calculation steps to determine physical properties or behaviors of these molecules.
Understanding the diatomic nature of a molecule helps in calculating properties such as molar mass, especially when analyzing crystal structures like the body-centred cubic (bcc) structure. The internal arrangement of atoms in a bcc lattice, where each unit cell effectively contains two atoms, aligns well with the diatomic configuration that allows specific calculation steps to determine physical properties or behaviors of these molecules.
Unit Cell Volume
The volume of a unit cell in crystal structures, like the body-centred cubic (bcc), directly relates to its geometric dimensions. In the case of the bcc lattice, you can determine the unit cell volume using the formula \(V = a^3\), where \(a\) is the cell edge length.
In this problem, the cell edge is given as \(300\text{ pm}\). First, we convert this from picometers to centimeters: \(300 \times 10^{-10} \text{ cm}\). The cube of the edge length, \(a\), gives the volume. Performing the calculations, \(V = (300 \times 10^{-10})^3\), results in \(27 \times 10^{-24} \text{ cm}^3\).
The calculation of unit cell volume is critical because it forms the basis for deriving mass and density properties of substances in solid-state physics. Understanding how to convert units and calculate volumes is essential in obtaining real-world measurements necessary for further molecular and material analysis.
In this problem, the cell edge is given as \(300\text{ pm}\). First, we convert this from picometers to centimeters: \(300 \times 10^{-10} \text{ cm}\). The cube of the edge length, \(a\), gives the volume. Performing the calculations, \(V = (300 \times 10^{-10})^3\), results in \(27 \times 10^{-24} \text{ cm}^3\).
The calculation of unit cell volume is critical because it forms the basis for deriving mass and density properties of substances in solid-state physics. Understanding how to convert units and calculate volumes is essential in obtaining real-world measurements necessary for further molecular and material analysis.
Molar Mass Calculation
To find the molar mass for a substance, especially in the context of the body-centred cubic (bcc) structure of a diatomic molecule like \(X_2\), we begin by determining the mass of the theoretical unit cell. Using the formula \(\text{Mass} = \text{Density} \times \text{Volume}\), substituting the density as \(6.17 \text{ g/cm}^3\) and the volume as calculated \(27 \times 10^{-24} \text{ cm}^3\), we find \(\text{Mass} = 1.6659 \times 10^{-22} \text{ g}\).
This mass represents two atoms within the unit cell, thus \(2m = 1.6659 \times 10^{-22} \text{ g}\). Solving for \(m\), the mass of one molecule of \(X_2\), yields \(m = 0.83295 \times 10^{-22} \text{ g}\). Converting this to g/mol using Avogadro's number \(N_A = 6 \times 10^{23}\), we get \(M = 49.977 \text{ g/mol}\).
Calculating molar mass is a foundational chemistry concept that helps in understanding the scale and mass of chemical substances in relation to their molecular count, critical for laboratory work and theoretical predictions.
This mass represents two atoms within the unit cell, thus \(2m = 1.6659 \times 10^{-22} \text{ g}\). Solving for \(m\), the mass of one molecule of \(X_2\), yields \(m = 0.83295 \times 10^{-22} \text{ g}\). Converting this to g/mol using Avogadro's number \(N_A = 6 \times 10^{23}\), we get \(M = 49.977 \text{ g/mol}\).
Calculating molar mass is a foundational chemistry concept that helps in understanding the scale and mass of chemical substances in relation to their molecular count, critical for laboratory work and theoretical predictions.
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