One atom of an element \(x\) weigh \(6.643 \times 10^{-23} \mathrm{~g}\). Number of moles of atom in \(20 \mathrm{~kg}\) is: (a) 4 (b) 40 (c) 100 (d) 500

Short Answer

Expert verified
The number of moles of atom in 20 kg of the element is 100 moles, which corresponds to option (c).

Step by step solution

01

Convert the Weight of Substance to Grams

To work with the number of moles, first convert the weight of the element from kilograms to grams. Since there are 1000 grams in a kilogram, multiply the weight of the substance by 1000.
02

Calculate the Number of Atoms

Divide the total weight of the element in grams by the weight of a single atom (also given in grams) to find the total number of atoms in the substance.
03

Use Avogadro's Number to Find Moles

One mole of a substance contains Avogadro's number of particles (atoms, molecules, etc.), which is approximately \(6.022 \times 10^{23}\). Divide the total number of atoms by Avogadro's number to find the number of moles.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Avogadro's Number
When studying chemistry, you'll frequently encounter Avogadro's number, which is an essential cornerstone of the mole concept. Avogadro's number, designated as approximately \(6.022 \times 10^{23}\), represents the number of particles, such as atoms or molecules, in one mole of a substance. This constant provides a link between microscopic substances at an atomic or molecular level and measurable quantities at a bulk scale.

Using Avogadro's number helps chemists convert between the number of particles and the number of moles. For example, if we say that we have 1 mole of carbon atoms, we are referring to approximately \(6.022 \times 10^{23}\) carbon atoms. This constant allows for calculations across the incredibly vast range of scales that chemists work with, from individual atoms to grams of material.
Molar Mass Calculation
Molar mass calculates how much one mole of a given substance weighs and is expressed in grams per mole (g/mol). It's found by summing the atomic masses of all the atoms in a molecule. The atomic mass of each element is listed on the periodic table and is measured in atomic mass units (amu).

To calculate the molar mass, simply identify each atom present in the molecule, count the number of each type, and multiply by the atomic mass from the periodic table. For elements, the molar mass is the same as the atomic mass listed on the periodic table. By determining the molar mass, we can convert between the mass of a substance and the number of moles, providing critical information for stoichiometry.
Converting Mass to Moles
To bridge the gap between the laboratory scale and the atomic scale, chemists use the concept of molar mass to convert the mass of a substance to the number of moles. The formula to do this is simple: by dividing the mass of the substance in grams by its molar mass, you obtain the number of moles.

For instance, let's consider a sample weighed in grams. By using the formula \( \text{Number of Moles} = \frac{\text{Mass of Substance (g)}}{\text{Molar Mass (g/mol)}} \), where the molar mass is taken from the periodic table or calculated, we can determine how many moles the sample contains. This calculation is a basic practical application of stoichiometry and one of the first skills mastered in chemistry.
Stoichiometry
Stoichiometry is a section within chemistry that involves the calculation of reactants and products in chemical reactions. It's based on the conservation of mass and the concept of moles. By using the balanced chemical equation and the known quantities of substances, stoichiometry allows you to predict how much product will form or how much reactant is necessary to create a certain amount of product.

Understanding stoichiometry requires a firm grasp of the mole concept, Avogadro's number, and molar mass calculations, as these all factor into defining the proportions of substances within a reaction. Through stoichiometric calculations, chemists can precisely scale chemical reactions from a laboratory bench to industrial production levels.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(25.0 \mathrm{~mm} \times 40: 0 \mathrm{~mm}\) piece of gold foil is \(0.25 \mathrm{~mm}\) thick. The density of gold is \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\). How many gold atoms are in the sheet? (Atomic weight : \(\mathrm{Au}=197.0\) ) (a) \(7.7 \times 10^{23}\) (b) \(1.5 \times 10^{23}\) (c) \(4.3 \times 10^{21}\) (d) \(1.47 \times 10^{22}\)

\(\mathrm{KMnO}_{4}\) reacts with oxalic acid according to the equation $$ 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O} $$ Here, \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KMnO}_{4}\) is equivalent to : (a) \(120 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (b) \(150 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (c) \(25 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (d) \(50 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Fe shows an oxidation state of \(+1\) in: (a) \(\mathrm{Fe}(\mathrm{CO})_{5}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right] \mathrm{SO}_{4}\) (c) \(\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}\) (d) \(\mathrm{FeCl}_{4}^{-}\)

In an organic compound of molar mass \(108 \mathrm{gm} \mathrm{mol}^{-1} \mathrm{C}, \mathrm{H}\) and \(\mathrm{N}\) atoms are present in \(9: 1:\) 3.5 by weight. Molecular formula can be : (a) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}\) (b) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{~N}\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{~N}_{3}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{18} \mathrm{~N}_{3}\)

A certain public water supply contains \(0.10 \mathrm{ppb}\) (part per billion) of chloroform \(\left(\mathrm{CHCl}_{3}\right) .\) How many molecules of \(\mathrm{CHCl}_{3}\) would be obtained in \(0.478 \mathrm{~mL}\) drop of this water ? (assumed \(d=1 \mathrm{~g} / \mathrm{mL}\) ) (a) \(4 \times 10^{-3} \times N_{A}\) (b) \(10^{-3} \times N_{A}\) (c) \(4 \times 10^{-10} \times N_{A}\) (d) None of these

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free