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Chapter 1: Problem 19

How many moles of electron weigh one kilogram? (a) \(\frac{1}{206}\) (b) \(\frac{1}{9.108} \times 10^{31}\) (c) \(\frac{6.023}{9.108} \times 10^{54}\) (d) \(\frac{1}{9.108 \times 6.023} \times 10^{8}\)

Short Answer

Expert verified
(d) \(\frac{1}{9.108 \times 6.023} \times 10^{8}\)

Step by step solution

01

Determine the Weight of One Mole of Electrons

The weight of one mole of electrons refers to the molar mass of electrons. The mass of an electron is approximately \(9.108 \times 10^{-31}\) kilograms. Therefore, the molar mass of electrons is \(9.108 \times 10^{-31} \times 6.023 \times 10^{23} \ \text{kg/mol}\) (using Avogadro's number, \(6.023 \times 10^{23}\) particles per mole).
02

Calculate Moles in One Kilogram

To find the number of moles in one kilogram, calculate the inverse of the molar mass of electrons: \[\text{Number of moles} = \frac{1}{9.108 \times 10^{-31} \times 6.023 \times 10^{23}}\] This simplifies to \[\frac{1}{(9.108 \times 6.023) \times 10^{-8}}\]
03

Simplify the Expression

The expression \(\frac{1}{(9.108 \times 6.023) \times 10^{-8}}\) can be simplified to \(\frac{1}{54.84984 \times 10^{-8}}\). Further simplification yields \(\frac{1}{54.84984} \times 10^{8}\), which closely matches option (d) \(\frac{1}{9.108 \times 6.023} \times 10^{8}\).

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Key Concepts

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

Molar Mass of Electrons
The molar mass of electrons is an important concept in chemistry that allows us to understand how much one mole of electrons would weigh. To find it, we need to multiply the mass of a single electron by Avogadro's number, which gives us the mass of one mole of electrons.

The mass of a single electron is approximately \(9.108 \times 10^{-31}\) kilograms. When this value is multiplied by Avogadro's number, \(6.023 \times 10^{23}\), it results in the molar mass of electrons, which gives us a value in kilograms per mole.
  • This calculation helps us understand the relationship between micro-scale particle quantities and the more tangible, macro-scale quantities.
  • The result is a crucial step in determining how many moles of electrons equate to a certain mass, such as in our problem about how many moles of electrons weigh one kilogram.
Avogadro's Number
Avogadro's number is a fundamental constant in chemistry, denoted as \(6.023 \times 10^{23}\) particles per mole. It is used to bridge the gap between the atomic scale and the macroscopic scale.

Whenever we talk about a mole of any substance, we are referring to Avogadro's number of those entities (atoms, molecules, electrons, etc.). For electrons, this means that one mole contains \(6.023 \times 10^{23}\) electrons.
  • This number allows for the calculation of atomic or molecular weight to a measurable amount.
  • It's essential for converting between atoms/molecules and mass, making it a cornerstone for molecular calculations.
Understanding Avogadro's number is critical for comprehending molecular calculations such as those in the study exercise.
Electron Mass
The electron mass is the weight of a single electron and is a key value in calculations involving electrons. Precisely, an electron has a mass of about \(9.108 \times 10^{-31}\) kilograms. This seemingly tiny number represents the weight of an individual electron, which is incredibly small due to the minuscule size of electrons themselves.

Knowing the mass of an electron is crucial for any calculations involving electron quantities, such as determining how many moles make up a certain weight.
  • This knowledge is essential in practical applications like determining the weight of a mole of electrons.
  • When multiplied by Avogadro's number, it gives us the molar mass, thereby linking subatomic particles to measurable quantities.
Mastering the concept of electron mass helps in understanding how individual atomic components contribute to overall mass.
Molecular Calculations
Molecular calculations enable us to understand and quantify the relationships between different molecular weights and quantities. At the core of these calculations is the ability to convert between mass and moles, thereby calculating how much of a substance is present.

In this exercise, we were asked to determine how many moles of electrons weigh one kilogram. By calculating the molar mass of electrons and dividing one kilogram by this molar mass, we find the number of moles.
  • Such calculations often involve multiplying and dividing by scientific constants like Avogadro's number and elemental or molecular masses.
  • These calculations are essential for practical chemical measurements and reactions in the laboratory.
Understanding molecular calculations helps students grasp how to measure and predict the behavior of substances in different chemical contexts.

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Most popular questions from this chapter

What weight of \(\mathrm{AgCl}\) will be precipitated when a solution containing \(4.77 \mathrm{~g}\) of \(\mathrm{NaCl}\) is added to a solution of \(5.77 \mathrm{~g}\) of \(\mathrm{AgNO}_{3} ?\) [1978]

The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) to form a stable coordination compound. Assume that both the reactions are \(100 \%\) complete. If \(1584 \mathrm{~g}\) of ammonium sulphate and \(952 \mathrm{~g}\) of \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is(Atomic weights in \(\mathrm{g} \mathrm{mol}^{-1}\) : \(\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{Cl}=35.5, \mathrm{Ca}=\) \(40, \mathrm{Ni}=59\) )

A sample of a hydrate of barium chloride weighing \(61 \mathrm{~g}\) was heated until all the water of hydration is removed. The dried sample weighed \(52 \mathrm{~g}\). The formula of the hydrated salt is : (atomic mass, \(\mathrm{Ba}=137\) \(\mathrm{amu}, \mathrm{Cl}=35.5 \mathrm{amu}\) ) [Main Online April 10,2015] (a) \(\mathrm{BaCl}_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{BaCl}_{2} .3 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{BaCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Dissolving \(120 \mathrm{~g}\) of a compound of (mol. wt. 60) in \(1000 \mathrm{~g}\) of water gave a solution of density \(1.12 \mathrm{~g} / \mathrm{mL}\). The molarity of the solution is: (a) \(1.00 \mathrm{M}\) (b) \(2.00 \mathrm{M}\) (c) \(2.50 \mathrm{M}\) (d) \(4.00 \mathrm{M}\)

An unknown chlorohydrocarbon has \(3.55 \%\) of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in \(1 \mathrm{~g}\) of chlorohydrocarbon are: (Atomic wt. of \(\mathrm{Cl}=35.5 \mathrm{u}\); Avogadro constant \(=6.023 \times 10^{23} \mathrm{~mol}^{-1}\) (a) \(6.023 \times 10^{9}\) (b) \(6.023 \times 10^{23}\) (c) \(6.023 \times 10^{21}\) (d) \(6.023 \times 10^{20}\)
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