Problem 118 - General Chemistry: Principles and Modern Applications

Question

Which answer is correct? One mole of liquid bromine, \(\mathrm{Br}_{2},\) (a) has a mass of \(79.9 \mathrm{g} ;\) (b) contains \(6.022 \times 10^{23}\) Br atoms; (c) contains the same number of atoms as in \(12.01 \mathrm{g} \mathrm{H}_{2} \mathrm{O} ;\) (d) has twice the mass of 0.500 mole of gaseous \(\mathrm{Cl}_{2}\)

Step-by-Step Solution

Answer

The correct statement is (c) One mole of liquid bromine, \(\mathrm{Br}_{2}\), contains the same number of atoms as in \(12.01 \mathrm{g}\) \(\mathrm{H}_{2} \mathrm{O}\).

Step 1: Validate Statement (a)

One mole of any element or compound is defined as its molar mass in grams. The molar mass of Bromine (\(\mathrm{Br}\)) is approximately 79.9 g/mol. However, liquid Bromine is \(\mathrm{Br}_{2}\), so its molar mass is \(79.9 \times 2 = 159.8 \mathrm{g/mol}\). Thus, one mole of liquid Bromine (\(\mathrm{Br}_{2}\)) does not have a mass of 79.9g. Hence statement (a) is incorrect.

Step 2: Validate Statement (b)

Avogadro's number (\(6.022 \times 10^{23}\)) states the number of atoms in exactly one mole of any substance. However, a mole of liquid Bromine is \(\mathrm{Br}_{2}\), which implies that each molecule consists of two Br atoms. Therefore, one mole of liquid Bromine (\(\mathrm{Br}_{2}\)) contains \(2 \times 6.022 \times 10^{23}\) Br atoms. Consequently, statement (b) is incorrect.

Step 3: Validate Statement (c)

This statement calls for an understanding of Avogadro's Law, which dictates that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. While the substances may differ, the number of molecules in a mole of any substance remains the same, that number being Avogadro's number (\(6.022 \times 10^{23}\)). Thus, one mole of \(\mathrm{Br}_{2}\) (Bromine) contains the same number of molecules as one mole of \(\mathrm{H}_{2} \mathrm{O}\) (Water). This makes statement (c) correct.

Step 4: Validate Statement (d)

We again refer to the molar mass for this verification. The molar mass of Chlorine (\(\mathrm{Cl}\)) is approximately 35.45 g/mol, which makes the molar mass of gaseous Chlorine (\(\mathrm{Cl}_{2}\)) about \(35.45 \times 2 = 70.9 \mathrm{g/mol}\). Therefore, 0.500 mole of gaseous Chlorine (\(\mathrm{Cl}_{2}\)) would have a mass of \(0.500 \times 70.9 = 35.45 \mathrm{g}\), which is not half the mass of one mole of liquid Bromine (\(159.8 \mathrm{g}\)). Thus, statement (d) is incorrect.

Key Concepts

Avogadro's Number

Avogadro's number, which is approximately \(6.022 \times 10^{23}\), represents a fundamental constant in chemistry. It defines the number of constituent particles, typically atoms or molecules, contained in one mole of a given substance.
This number is named after the scientist Amedeo Avogadro, who was instrumental in making atomic and molecular concepts measurable.
Understanding Avogadro's number helps translate macroscopic measurements to the atomic level, which is crucial in many scientific calculations.

• It's used to calculate how many atoms or molecules are in a macroscopic sample.
• It applies to all substances, regardless of their form, homogeneous or heterogeneous.

In this specific exercise, Avogadro's number allows us to calculate that one mole of liquid bromine \(\mathrm{Br}_{2}\) actually contains \(2 \times 6.022 \times 10^{23}\) atoms, due to each \(\mathrm{Br}_{2}\) molecule having two bromine atoms. This concept highlights how a unique atomic scale count (like atoms per molecule) still maintains the overall molecule-based count defined by Avogadro's number.

Molecular Composition

The molecular composition of a compound refers to the arrangement and types of atoms present in its molecules.
For instance, liquid bromine is composed of diatomic molecules, denoted as \(\mathrm{Br}_{2}\). This notation informs us directly that each molecule is made up of two bromine (Br) atoms.
Understanding the molecular composition is crucial because it helps determine the molar mass of the substance.

• Molar mass is calculated by adding up the atomic masses of all the atoms in a molecule.
• This determines how much one mole of a particular molecule weighs.

In the exercise, the molar mass of \(\mathrm{Br}_{2}\) is not simply double the molar mass of a single Br atom. Depending on the element's molecular configuration, the total mass needs to account for the molecular formulation, thus \(79.9 \times 2 = 159.8 \) grams for \(\mathrm{Br}_{2}\). Understanding this helps clarify why some statements in the exercise are incorrect.

Chemical Calculations

Chemical calculations involve converting between moles, grams, and the number of molecules or atoms, often using Avogadro's number and molar mass. They form the backbone for predicting how substances will behave chemically and physically in a given situation.
By applying these calculations, you can determine the amount of substance in a reaction, its composition, or how it will interact with other chemicals.

• Start with known values: molar masses from the periodic table, given mass in grams, or moles.
• Use conversion factors such as molar mass or Avogadro’s number to shift between units.

In the exercise, such calculations are used to compare the mass and quantity of \(\mathrm{Br}_{2}\) and \(\mathrm{Cl}_{2}\). Properly understanding chemical calculations can prevent errors and enhance comprehension in evaluating the accuracy of provided statements. Ensuring a thorough grasp of these concepts is critical for a proficient chemical analysis.