(C) The average atomic mass is calculated using the formula:$A_{avg} = \frac{\sum (A_i \times X_i)}{\sum X_i}$Given:$A_1 = 10, X_1 = 20\% = 0.2$$A_2 =

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(C) The average atomic mass is calculated using the formula:$A_{avg} = \frac{\sum (A_i \times X_i)}{\sum X_i}$Given:$A_1 = 10, X_1 = 20\% = 0.2$$A_2 = 11, X_2 = 80\% = 0.8$Substituting the values:$A_{avg} = \frac{(10 \times 0.2) + (11 \times 0.8)}{0.2 + 0.8}$$A_{avg} = \frac{2.0 + 8.8}{1.0}$$A_{avg} = 10.8$Thus,the atomic weight of boron is $10.80$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

Medium

Naturally occurring boron has $20\%\, _5B^{10}$ and $80\%\, _5B^{11}$. The atomic weight of boron is :-

A
$10.50$
B
$11.0$
C
$10.80$
D
$10.20$

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

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

Naturally occurring boron has $20\%\, _5B^{10}$ and $80\%\, _5B^{11}$. The atomic weight of boron is :-

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Calculate the molar mass of naturally occurring argon from the data given in the following table: Isotope Molar Mass $(g \ mol^{-1})$ Abundance $^{36}Ar$ $35.96755$ $0.337 \%$ $^{38}Ar$ $37.96272$ $0.063 \%$ $^{40}Ar$ $39.9624$ $99.600 \%$

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Calculate the molar mass of naturally occurring argon from the data given in the following table:

Isotope Molar Mass $(g \ mol^{-1})$ Abundance

$^{36}Ar$ $35.96755$ $0.337 \%$

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

$^{40}Ar$ $39.9624$ $99.600 \%$