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 3 of 3: Perrin sedimentation method

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m^* = \tfrac{4}{3}\pi r^3(\rho_p - \rho_w) = 8.3 \times 10^{-18}\ \text{kg};\quad \text{slope} = -\dfrac{N_A m^* g}{RT} \Rightarrow N_A = \dfrac{RT \times 0.0235 \times 10^{6}}{m^* g} \approx 6.9 \times 10^{23}\ \text{mol}^{-1}

Analysis

The particle mass is $\tfrac{4}{3}\pi r^3 \rho_p = 4.81 \times 10^{-17}$ kg; subtracting buoyancy gives $m^* = 8.3 \times 10^{-18}$ kg. In $\ln(n_h/n_{h_0}) = -N_A m^* g (h-h_0)/RT$ the slope is $-0.0235\ \mu\text{m}^{-1} = -2.35 \times 10^{4}\ \text{m}^{-1}$ , so $N_A = RT \times 2.35 \times 10^{4}/(m^* g) \approx 6.9 \times 10^{23}$ mol $^{-1}$ .

Common pitfall. Use the effective (buoyancy-corrected) mass and convert the slope from

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

\text{m}^{-1}