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Question

The face-centered unit cell of nickel has an edge length of $$352.39\ pm$$. The density of nickel is $$8.9\ g\ cm^{-3}$$. Calculate the value of Avogadro's number. The atomic mass of nickel is $$58.7$$ and $$1\ pm$$ is equal to $$10^{-10} cm$$.


Solution

$$a=352.39pm$$        $$\rho=8.9g/cm^3$$, $$M_A=58.7$$, $$N_A=?$$ , $$Z=4$$ FCC

$$\rho=\cfrac {Z\times M_A}{N_A\times a^3}$$
$$8.9=\cfrac {4\times 58.7}{N_A\times 43.76\times 10^{-24}}\Rightarrow N_A=6.0287\times 10^{23}$$

Chemistry

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