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A small cube of iron is observed under a microscope. The edge of the cube is $4.00 \times 10^{-6}$ cm long. The atomic mass of iron is 55.9 u, and its density is 7.86 $g/cm^{3}$.

a) Find the mass of the cube.

b) Find the number of iron atoms in the cube.

My work:

$d = \frac{m}{v} \implies m = d \times v$

$m = (6.4 \times 10^{-17} cm^3) \times (7.86 \frac{g}{cm^3})$

$m = 5.0304 \times 10^{-16}$ g

Is this correct? Also how do I find part b?

user45417
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  • Looks right to me. For b. you simply need to conver the atomic mass of iron to mass in grams, and devide it. That'll give you the number of iron atoms needed to reach that mass. – Studentmath Jun 01 '14 at 16:25

1 Answers1

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Your answer to part $(a)$ is correct. For part $(b)$, find the number of moles of iron available with you first:

$$\text{No. of moles}=\dfrac{\text {weight of substance in grams}}{\text{molecular mass of substance in amu}}$$

Hence in the no. of moles($n$) is: $$n=\dfrac{5.0304×10 ^{−16} \text{ g}}{55.9 \text { amu}}$$

$$\implies n\approx 9.0054 \times 10^{-18}$$

$$\implies \text {no. of iron atoms in the cube}=\text {no. of moles}\times \text{Avogadro's number}$$

$$\implies\text {no. of iron atoms in the cube}=n\times 6.023\times 10^{23}$$

$$\implies\text {no. of iron atoms in the cube}\approx 5.424 \times 10^6$$

Apurv
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