Avogadro's number is a fundamental constant in chemistry. However, an accurate determination of this — Physical Chemistry — Kinetics Chemistry Question
Physical methods for determination of Avogadro’s number
Avogadro's number is a fundamental constant in chemistry. However, an accurate determination of this value took a long time. Avogadro (1776–1856) himself did not know Avogadro's number as it is known today. At about the time of his death, Avogadro's number determined from gas properties, such as diffusion coefficient and viscosity, approached 5×1022. Avogadro's number as we know it today (6.02×1023) became available only in the early 20th century. Let's consider three separate approaches.
At thermal equilibrium, the probability of finding a molecule with a mass m at height h is proportional to the Boltzmann factor, exp(–E(h) / kBT), where E(h) is the gravitational potential energy (mgh, where g is 9.81 m s–2) and kB is the Boltzmann constant. Thus, the number density at h follows "barometric" distribution:
(a) Spherical particles of diameter 0.5 µm and density 1.10 g cm–3 are suspended in water (density 1.00 g cm–3) at 20 °C. Calculate the effective mass m of the particles corrected for buoyancy.
(b) Now the number density of the particles with effective mass will follow barometric distribution. In an experiment where a vertical distribution of such particles was measured, it was observed that the number density at h decreased to 1/e times the number density at ho over a vertical distance of 6.40×10–3 cm. Calculate Boltzmann’s constant.
(c) Calculate Avogadro's number using Boltzmann’s constant and the gas constant. (R = 8.314 J mol–1K–1)
Model Answer
(a) Volume of the particle = (4/ 3 × 3.14) (0.5×10–6 / 2)3 m3 = 6.54×10–14 cm3
Effective mass = (6.54×10–14 cm3 ) (1.10 – 1.00) g cm–3 = 6.54×10–15 g
(b) = 1
kB = 6.54×10–15 kg × 9.81m s–2 × 6.40×10–5 m / 293.15 K = 1.40×10–23 J K–1
(c) Avogadro's number = R / kB = 8.314 J mol–1K–1 / 1.40×10–23 J K–1 = 5.94×1023 mol–1
Avogadro's number can also be determined by single crystal X–ray crystallography. The density of sodium chloride crystal is 2.165 g cm–3. The sodium chloride lattice is shown below (Figure 8.1).
Figure 8–1. Lattice structure of sodium chloride
In the rock–salt structure one finds a face–centered cubic array of anions and the same array of cations. The two arrays interpenetrate each other. A unit cell contains 4 anions (8 centered at the apexes are each shared by 8 unit cells thus giving 1 anion, and 6 positioned at the face centers are each shared by 2 unit cells giving 3 anions). A unit cell also contains 4 cations.
The distance between the centers of adjacent Na+ and Cl– ions was determined to be 2.819×10–8 cm. Calculate the Avogadro's number.
Model Answer
Length of the edge of the unit cell = 2 × 2.819×10–8 cm = 5.638×10–8 cm
Volume of the unit cell = (5.638×10–8 cm)3 = 1.792 × 10–22 cm3
Volume per Na+ plus Cl– = 1.792×10–22 cm3 / 4 = 4.480×10–23 cm3
Formula mass of NaCl = 22.99 + 35.45 = 58.44
Molar volume of the crystal = 58.44 g / 2.165 g cm–3 = 26.99 cm3
Avogadro's number = 26.99 cm3 / 4.480×10–23 cm3 = 6.025×1023
In a well known oil drop experiment, Millikan determined in 1913 that the basic unit of electric charge is 1.593×10–19 C. Calculate Avogadro's number from this value and Faraday charge, which is electric charge per equivalent (1 Faraday = 96,496 coulomb as used by Millikan).
Model Answer
Avogadro’s number = 96496 C mol–1 / 1.593×10–19 C = 6.058×1023 mol–1