Question

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Answer 1

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Answer 2

Answer:

$\bigstar$ Given:

$\implies \rm ^2^0^0\;X:90\%$

$\implies \rm ^1^9^9\;X:8.0\%$

$\implies \rm ^2^0^2\;X:2.0\%$

$\bigstar$ To FInd:

⇒ Average atomic mass of the naturally-occurring element X

$\bigstar$ Solution:

⇒ A = Isotopic mass  

⇒ X = Percentage of occurrence

Where, (From given data)

$\sf A_1 = 200\\A_2 = 199\\A_3 = 202$

$\sf X_1 = 90\\X_2 = 8\\X_3 = 2$

The Formula to be used here is,

$\underline{\large {\rm Average\;atomic\;mass = k }}$

$\boxed{\rm k = \dfrac{(A_1 \times X_1 )+ (A_2 \times X_2)+ (A_3 \times X_3)}{3 \times 100}}$

By substituting the above values in the formula

We get,

$\rm k = \dfrac{(200 \times 90)+ (199 \times 8)+ (202 \times 2)}{3 \times 100}$

After Solving,

$\red{\boxed{\boxed{\rm {Average\; atomic\; mass=199.96}}}}$