Calculate the atomic mass (average) of chlori | vedclass

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(D) The average atomic mass is calculated by the weighted average of the isotopes: $	ext{Average atomic mass} = \frac{(	ext{Abundance of } ^{35}Cl \

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$35.45$

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(D) The average atomic mass is calculated by the weighted average of the isotopes: $	ext{Average atomic mass} = \frac{(	ext{Abundance of } ^{35}Cl 	imes 	ext{Mass of } ^{35}Cl) + (	ext{Abundance of } ^{37}Cl 	imes 	ext{Mass of } ^{37}Cl)}{100}$ $	ext{Average atomic mass} = \frac{(75.77 	imes 34.9689) + (24.23 	imes 36.9659)}{100}$ $	ext{Average atomic mass} = \frac{2649.59 + 895.68}{100} = \frac{3545.27}{100} = 35.4527 \ u$ Rounding to two decimal places,we get $35.45 \ u$.

Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

Column $-I$	Column $-II$
$(i)$ Barium sulphate	$(a)$ $1877 \ g/mol$
$(ii)$ Magnesium pyrophosphate	$(b)$ $235 \ g/mol$
$(iii)$ Ammonium phosphomolybdate	$(c)$ $188 \ g/mol$
$(iv)$ Silver chloride	$(d)$ $222 \ g/mol$
$(v)$ Silver bromide	$(e)$ $143.5 \ g/mol$
$(vi)$ Silver iodide	$(f)$ $233 \ g/mol$

Calculate the atomic mass (average) of chlorine using the following data:

Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

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Calculate the atomic mass (average) of chlorine using the following data: Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$ $^{35}Cl$ $75.77$ $34.9689$ $^{37}Cl$ $24.23$ $36.9659$ (in $u$)