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(A) The average atomic mass is calculated by taking the weighted average of the masses of the isotopes based on their relative abundance.Formula: $	e

Vedclass · Accepted Answer

$55.95$

Vedclass · Answer

(A) The average atomic mass is calculated by taking the weighted average of the masses of the isotopes based on their relative abundance.Formula: $	ext{Average Atomic Mass} = \frac{(M_1 	imes P_1) + (M_2 	imes P_2) + (M_3 	imes P_3)}{100}$Given:Isotope $1$: Mass $= 54$,Abundance $= 5\%$Isotope $2$: Mass $= 56$,Abundance $= 90\%$Isotope $3$: Mass $= 57$,Abundance $= 5\%$Calculation:$	ext{Average Atomic Mass} = \frac{(54 	imes 5) + (56 	imes 90) + (57 	imes 5)}{100}$$= \frac{270 + 5040 + 285}{100}$$= \frac{5595}{100} = 55.95 \ u$

Given that a natural sample of iron $(Fe)$ has isotopes $_{26}^{54}Fe$,$_{26}^{56}Fe$,and $_{26}^{57}Fe$ in the ratios of $5\%$,$90\%$,and $5\%$ respectively. What will be the average atomic mass of iron $(Fe)$ (in $u$)?

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