Question

An element X has following isotopic composition 56X=90%, 57X=8% , 59X=2%. The weighted average atomic mass will be equal to

Answer 1

Average atomic mass = (%isotope1x mass of isotope1 + % isotope 2 x mad of isotope 2)÷ 100

so here avg atm mass = [(56x90) + (57x8) + ( 59x2)]÷100
=56.14 u

Answer 2

Answer:

56.14 amu

Explanation:

Isotopes are the atoms of an element that have same number of protons but different number of neutrons.

And The average atomic mass of any element is defined as the sum of the atomic masses of each isotope multiplied by their natural abundance.

Now, the given element has 3 isotopes, 56X, 57X and 59X.

Atomic mass of 56X = 56 amu

Natural abundance of 56X = 90% = 0.90

Atomic mass of 57X = 57 amu

Natural abundance of 57X = 8 % = 0.08

Atomic mass of 59X = 59 amu

Natural abundance of 59X = 2 % = 0.02

Therefore, the average atomic mass of element X is calculated as:

$Average\ atomic\ mass\ of\ X\ =( Atomic\ mass\ of\ 56X)$×$(abundance\ of\ 56X)$+$( Atomic\ mass\ of\ 57X)$×$(abundance\ of\ 57X)$+$( Atomic\ mass\ of\ 59X)$×$(abundance\ of\ 59X)$.

$Average\ atomic\ mass\ of\ X=$$(56$×$0.90)$ +$(57$×$0.08)$+$(59$×$0.02)$$= 56.14$ amu

Hence, the weighted average atomic mass = 56.14 amu

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