Problem 1

The Avogadro constant

N_A

can be determined by three independent methods. (Method A) X-ray diffraction shows that gold crystallises in a face-centred cubic unit cell of edge

a = 0.408

nm; the density of Au is

1.93 \times 10^{4}\ \text{kg m}^{-3}

and

M(\mathrm{Au}) = 196.97\ \text{g mol}^{-1}

. (Method B, Rutherford 1911) A purified sample of 192 mg of

\ce{^{226}Ra}

stands 40 days, after which its total

\alpha

-decay rate is 27.7 GBq; sealing it for a further 163 days produces 10.4

\text{mm}^3

of He measured at 101325 Pa and 273 K (

\alpha

particles become He atoms). (Method C, Perrin 1909) Colloidal spheres of radius

2.12 \times 10^{-7}

m and density

1.206 \times 10^{3}\ \text{kg m}^{-3}

suspended in water (

\rho = 999\ \text{kg m}^{-3}

) at 15 \°C show a Boltzmann height distribution: a plot of

\ln(n_h/n_{h_0})

versus

(h - h_0)

has slope

-0.0235\ \mu\text{m}^{-1}

. Determine

N_A

from each method.

Step 1 of 3: X-ray diffraction method

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N = 8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4\ \text{Au atoms per cell};\quad m_{\text{cell}} = \rho a^3 = 1.31 \times 10^{-24}\ \text{kg};\quad N_A = \dfrac{M}{m_{\text{cell}}/4} = 6.01 \times 10^{23}\ \text{mol}^{-1}

Analysis

An fcc cell contains $8 \times 1/8 + 6 \times 1/2 = 4$ atoms. Its mass is $\rho a^3 = 1.93 \times 10^{4} \times (0.408 \times 10^{-9})^3 = 1.31 \times 10^{-24}$ kg, so one Au atom weighs $3.28 \times 10^{-25}$ kg $= 3.28 \times 10^{-22}$ g and $N_A = 196.97/3.28 \times 10^{-22} = 6.01 \times 10^{23}$ mol $^{-1}$ .

Common pitfall. In an fcc cell, corner atoms count 1/8 and face atoms 1/2; do not count 8 + 6 = 14 atoms.