An element,X has the following isotopic compo | vedclass

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(D) The weighted average atomic mass is calculated as:$	ext{Average atomic mass} = (\% 	ext{ abundance of isotope } 1 	imes 	ext{mass of isotope }

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$200$

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(D) The weighted average atomic mass is calculated as:$	ext{Average atomic mass} = (\% 	ext{ abundance of isotope } 1 	imes 	ext{mass of isotope } 1) + (\% 	ext{ abundance of isotope } 2 	imes 	ext{mass of isotope } 2) + (\% 	ext{ abundance of isotope } 3 	imes 	ext{mass of isotope } 3)$$	ext{Average atomic mass} = (0.90 	imes 200) + (0.08 	imes 199) + (0.02 	imes 202)$$	ext{Average atomic mass} = 180.00 + 15.92 + 4.04 = 199.96 \ amu$Rounding to the nearest whole number,the value is $200 \ amu$.

Isotope	Molar Mass $(g \ mol^{-1})$	Abundance
$^{36}Ar$	$35.96755$	$0.337 \%$
$^{38}Ar$	$37.96272$	$0.063 \%$
$^{40}Ar$	$39.9624$	$99.600 \%$

An element,$X$ has the following isotopic composition:
$^{200}X: 90\%$,$^{199}X: 8.0\%$,$^{202}X: 2.0\%$
The weighted average atomic mass of the naturally occurring element $X$ is closest to......$amu$.

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$^{38}Ar$ $37.96272$ $0.063 \%$

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An element,$X$ has the following isotopic composition:$^{200}X: 90\%$,$^{199}X: 8.0\%$,$^{202}X: 2.0\%$The weighted average atomic mass of the naturally occurring element $X$ is closest to......$amu$.