Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Indium (atomic weight \(=114.82\) ) has two naturally occurring isotopes, the predominant one form has isotopic weight \(114.9041\) and abundance of \(95.72 \%\). Which of the following isotopic weights is the most likely for the other isotope ? (a) \(112.94\) (b) \(115.90\) (c) \(113.90\) (d) \(114.90\)

Short Answer

Expert verified
The most likely isotopic weight for the other isotope is (c) 113.90.

Step by step solution

01

Calculate the weighted average

Use the formula for average atomic mass: Average Atomic Mass = (Isotope 1 Mass * Isotope 1 Abundance) + (Isotope 2 Mass * Isotope 2 Abundance). Since the average atomic mass and the mass and abundance of Isotope 1 are given, we can solve for the mass of Isotope 2.
02

Determine the abundance of the second isotope

The abundance of the second isotope will be 100% - the abundance of the first isotope. This is because the abundances of all isotopes must add up to 100%.
03

Substitute known values into the formula

Substitute the known values into the average atomic mass formula and solve for the mass of Isotope 2.
04

Solve for the mass of Isotope 2

After substituting the values into the formula, solve the equation for the mass of Isotope 2. This will give you the most likely isotopic weight for the other isotope.

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.

Average Atomic Mass Calculation
Understanding average atomic mass calculation is fundamental in physical chemistry. The average atomic mass is not just a simple average of isotopic masses. Instead, it is a weighted average that factors in both the mass and the isotopic abundance of each isotope of an element.

To calculate the average atomic mass of an element, you need to use the following formula: \
\[\begin{equation} \text{Average Atomic Mass} = (\text{Isotope 1 Mass} \times \text{Isotope 1 Abundance}) + (\text{Isotope 2 Mass} \times \text{Isotope 2 Abundance}) + \ldots \end{equation}\]\
The abundances are expressed as decimal fractions. For example, an abundance of 95.72% would be 0.9572 in this formula. By multiplying the mass of each isotope by its respective abundance and summing these values, you obtain the average atomic weight for the element as listed in the periodic table.
Isotopic Abundance
The term isotopic abundance refers to the relative amount in which each isotope of an element appears naturally. For any given element, the sum of the abundances of its naturally occurring isotopes equals 100%. It's crucial to understand that the abundances affect the average atomic mass significantly because without considering it, the calculations would be inaccurate.

When knowing the abundance of one isotope and the average atomic mass, the abundance of another isotope can be deduced since all abundances must add to 100%. In the given exercise example, knowing one isotope's abundance at 95.72% directly tells us the second isotope must have an abundance of 4.28% (100% - 95.72%). This simplified example involves only two isotopes, but remember that elements with more isotopes would have a more complex calculation, considering the abundance of each.
Isotope Determination
The process of isotope determination involves identifying the mass of different isotopes and their abundances in a given sample. Knowing this information allows chemists to calculate the average atomic mass of an element and hence, determine the identity of unknown isotopes.

In order to solve for the mass of an unknown isotope when one isotope's mass and abundance are known, and the average atomic mass is given, the balance equation method is used. You set up an equation where the sum of the masses of the isotopes, each multiplied by their respective abundance, equals the average atomic mass. By inserting the known values and solving for the unknown, you can identify the most likely isotopic weight of the remaining isotope. This calculation is fundamental in fields such as radiochemistry, where isotope determination is essential for identifying radioisotopes.

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

Calculate the molality of \(1 \mathrm{~L}\) solution of \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}(w / V)\), given that the density of the solution is \(1.80 \mathrm{~g} \mathrm{~mL}^{-1}\). (a) \(8.16\) (b) \(8.6\) (c) \(1.02\) (d) \(10.8\)

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) \(\mathrm{FeCl}_{4}^{-}\)

Cisplatin, an anticancer drug, has the molecular formula \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2} .\) What is the mass (in gram) of one molecule ? (Atomic weights : \(\mathrm{Pt}=195, \mathrm{H}=1.0, \mathrm{~N}=14, \mathrm{Cl}=35.5\) ) (a) \(4.98 \times 10^{-21}\) (b) \(4.98 \times 10^{-22}\) (c) \(6.55 \times 10^{-21}\) (d) \(3.85 \times 10^{-22}\)

The oxidation number of phosphorus in \(\mathrm{Ba}\left(\mathrm{H}_{2} \mathrm{PO}_{2}\right)_{2}\) is : (a) \(-1\) (b) \(+1\) (c) \(+2\) (d) \(+3\)

In an iodometric estimation, the following reactions occur \(2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2} ; \mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{NaI}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) \(0.12\) mole of \(\mathrm{CuSO}_{4}\) was added to excess of \(\mathrm{KI}\) solution and the liberated iodine required \(120 \mathrm{~mL}\) of hypo. The molarity of hypo solution was: (a) 2 (b) \(0.20\) (c) \(0.1\) (d) \(1.0\)

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