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myrzilka [38]
3 years ago
9

g Indium has two naturally occurring isotopes: 113In with an atomic weight of 112.904 amu, and 115In with an atomic weight of 11

4.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-of-occurrences of 115In isotope. Round off the answer to three significant figures.
Biology
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

Fraction-of-occurrences of 115In isotope = 0.957

Explanation:

As we know  weight of 113 In  isotope is equal to the sum of product of fraction-of-occurrences of two constituents isotopes and their atomic weight

^{115}In_w = f_{113}*A_{113} + f_{115}*A_{115} ------(1)

As we know sum of fraction-of-occurrences of two constituents isotopes is equal to one.

f_{113} + f_{115} = 1\\f_{113} = 1- f_{115}---------(2)

Substituting the given values in equation 1, we get

^{115}In_w = f_{113}*A_{113} + f_{115}*A_{115}\\114.818 = (1 - f_{115}) * 112.904 + f_{115} * 114.904\\f_{115} = 0.957

Substituting this value in equation 2 we get -

f_{113} = 1- f_{115}\\f_{113} =  1- 0.957 \\f_{113} = 0.043

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