Question

g Indium has two naturally occurring isotopes: 113In with an atomic weight of 112.904 amu, and 115In with an atomic weight of 114.904 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.

Answer 1

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 -

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