An iron alloy contains 54.7% iron and has a d | vedclass

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(C) Volume of the block = $10 \, cm 	imes 15 \, cm 	imes 20 \, cm = 3000 \, cm^3$. Mass of the block = $	ext{Volume} 	imes 	ext{Density} = 3000 \

Vedclass · Accepted Answer

$14.41 	imes 10^{25}$

Vedclass · Answer

(C) Volume of the block = $10 \, cm 	imes 15 \, cm 	imes 20 \, cm = 3000 \, cm^3$. Mass of the block = $	ext{Volume} 	imes 	ext{Density} = 3000 \, cm^3 	imes 8.17 \, g \, cm^{-3} = 24510 \, g$. Mass of iron in the block = $\frac{54.7}{100} 	imes 24510 \, g = 13406.97 \, g$. Number of moles of $Fe = \frac{13406.97 \, g}{55.85 \, g \, mol^{-1}} \approx 240.05 \, mol$. Number of $Fe$ atoms = $	ext{moles} 	imes N_A = 240.05 	imes 6.022 	imes 10^{23} \approx 1.445 	imes 10^{26}$. Given the options provided,the closest calculated value is $14.41 	imes 10^{25}$.

An iron alloy contains $54.7\%$ iron and has a density of $8.17 \, g \, cm^{-3}$. What is the number of iron atoms in an alloy block of dimensions $10 \, cm \times 15 \, cm \times 20 \, cm$?

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