An element, X has the following isotopic composition, 200X:90% 199X:8% 202X:2%The weighted average atomic mass of the naturally occurring element ‘X’ is closest to _____.

Average atomic mass ofX=90100×200+8100×199+2100×202=199.96 amu ≈ 200 amu

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

An element, X has the following isotopic composition,
$\begin{array}{l} ^{200} X : 90 % \\ ^{199} X : 8 % \\ ^{202} X : 2 % \end{array}$
The weighted average atomic mass of the naturally occurring element ‘X’ is closest to _____.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Average atomic mass of

$X = [\frac{90}{100} \times 200] + [\frac{8}{100} \times 199] + [\frac{2}{100} \times 202]$

$= 199.96 amu \approx 200 amu$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE