indium(atomic wt.= 114.82 u ) has two naturally occuring isotopes, the predominant one from has isotopic wt. 114.9041 and abundance of 95.72%. which of the following isotopic wt. is the most likely for the other isotope?
Let atomic weight of other isotope is M
114.82= 114.9041×95.72 + 4.28×M ÷ 100
M = 112.94
no answer choices.
abundance first isotope = 95.72%
abundance other isotope is 100-95.72 = 4.28%
Let x = mass other isotope
(114.9041*0.9572) + (x*0.0428) = 114.82
Solve for x
112.94
112.94
To determine the isotopic weight of the other isotope of indium, we can use the information provided. Indium has two naturally occurring isotopes, with their respective isotopic weights and abundances.
Let's assume the isotopic weight of the other isotope, which has an abundance of less than 95.72%, is x. Using the given information, we can set up an equation:
(Weight of Isotope 1 * Abundance of Isotope 1) + (Weight of Isotope 2 * Abundance of Isotope 2) = Average Atomic Weight of Indium
Substituting the known values:
(114.9041 u * 95.72%) + (x * (100% - 95.72%)) = 114.82 u
The abundance of the other isotope can be expressed as (100% - 95.72%), or (4.28%). Solving the equation for x:
(114.9041 u * 95.72%) + (x * 4.28%) = 114.82 u
(109.985 u) + (0.0428x) = 114.82 u
To isolate x, we subtract 109.985 u from both sides:
(0.0428x) = 114.82 u - 109.985 u
(0.0428x) = 4.835 u
Now, divide both sides of the equation by 0.0428 to solve for x:
x = (4.835 u) / 0.0428
x ≈ 113.09 u
Therefore, the most likely isotopic weight for the other isotope of indium is approximately 113.09 u.