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Standard X

Chemistry

Atom

An clement, X...

Question

An clement, X has the following isotopic composition,
$^{299} X : 90$ %
$^{199} X : 8$ %
$^{202} X : 2$ %
the weighted average atomic mass of the naturally- occurring element 'X' is closest to:

201 amu

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202 amu

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199 amu

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200 amu

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Solution

The correct option is D 200 amu
Average atomic mass = $Σ$ percentage x atomic mass of isotope
= $\frac{90 \times 200 + 8 \times 199 + 2 \times 202}{100}$
=200 amu

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An clement, X has the following isotopic composition, 299X:90% 199X:8% 202X:2% the weighted average atomic mass of the naturally- occurring element 'X' is closest to:

The correct option is D 200 amuAverage atomic mass = Σ percentage x atomic mass of isotope = 90×200+8×199+2×202100 =200 amu

An clement, X has the following isotopic composition,
$^{299} X : 90$ %
$^{199} X : 8$ %
$^{202} X : 2$ %
the weighted average atomic mass of the naturally- occurring element 'X' is closest to:

The correct option is D 200 amu
Average atomic mass = $Σ$ percentage x atomic mass of isotope
= $\frac{90 \times 200 + 8 \times 199 + 2 \times 202}{100}$
=200 amu