Question

Avogadro's number can be computed from the following measured properties of pure crystalline silicon: ${\mathbf{}}^{\mathbf{5}}$

(1) atomic mass (obtained from the mass and abundance of each isotope), (2) density of the crystal, (3) size of the unit cell (the smallest repeating unit in the crystal), and (4) number of atoms in the unit cell. For the material that was used, the average atomic mass of Si is role="math" localid="1663313481329" ${\mathbf{m}}_{\mathbf{si}}\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{085}\mathbf{}\mathbf{384}\mathbf{}\mathbf{2}\left(35\right)\mathbf{g}\mathbf{}\mathbf{/}\mathbf{}\mathbf{mol}$, where 35 is the uncertainty (standard deviation) in the last two digits. The density is $\mathbf{p}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{329}\mathbf{}\mathbf{031}\mathbf{}\mathbf{9}\mathbf{}\mathbf{\left(}\mathbf{18}\mathbf{\right)}\mathbf{g}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, the size of the cubic unit cell is ${\mathbf{c}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{431}\mathbf{}\mathbf{020}\mathbf{}\mathbf{36}\mathbf{}\left(33\right)\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{cm}$, and there are 8 atoms per unit cell. Avogadro's number is computed from the equation

${\mathbf{N}}_{\mathbf{A}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{Si}}}{\left({\mathrm{pc}}_0^3\right)\mathbf{/}\mathbf{8}}$

From the measured properties and their uncertainties, compute Avogadro's number and its uncertainty.

To find the uncertainty of ${\mathbf{c}}_{\mathit{o}\mathit{n}}^{\mathbf{3}}$ use the function $\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{4}}$in Table 3-1.

Short answer

The uncertainty =$4.8\times {10}^{17}$

Step-by-step solution

Given

⦁ The Average atomic mass of Si is $\left(m_{\mathrm{Si}}\right)\mathbf{=}\mathbf{28}\mathbf{.}\mathbf{0853842}\left(\pm 35\right)\mathbf{g}\mathbf{/}\mathbf{mol}$

where: 35 is the uncertainty (standard deviation)

⦁ Density (p)=2.3290319localid="1663314180435" $\left(\pm 18\right)\mathbf{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$

⦁ The size of the cubic unit cell is $\left(c_o\right)\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{43102036}\left(\pm 33\right)\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$

⦁ 8 atoms per unit cell

Substituting values to find the value of uncertainty

$$\begin{gathered} N_A=\frac{m_{si}}{{\displaystyle \frac{\left(p\right)\left(c_0^3\right)}{8}}} \\ {N}_A=\frac{28.0853842\left(\pm 35\right)g/mol}{\left(2.3290319\left(\pm 18\right)g/c{m}^3\right)\left(5.43102036\left(\pm 33\right)\times {10}^{-8}\right)} \\ {N}_A=6.0221369\times {10}^{23} \\ Uncertainty=4.8\times {10}^{17} \end{gathered}$$