Question

Given below are data on a gum-resin suspension in water at 22°C. Average radius of a grain of suspension = 0.2 μm (1 μm = 10^-6 m), average mass of grain = 6.2 x 10^-17 kg, average concentration in a layer = 43 particles pre unit area at some reference level, average concentration in a layer = 100.3 particles per unit area at a level 11 μm lower. Estimate the value of Avogadro's number from this data. Compare your answer with the correct value.

Answer 1

Here, T = 22°C = 273 + 22 = 295 K

Average radius of grain, r = 0.2 μm = 0.2 x 10^-6 m

Volume (V) = 4/3 πr³ = 4/3 x 3.14 x (0.2 x 10^-6)³

= 3.35 x 10^-20 m³

Density of grain, ρ = m/V = {6.2 x 10^-7}/{3.35 x 10^-20} = 1851 kg m^-3

n₁ = 43 m^-2

and n₂ = 100.3 m^-2

h₂ - h₂ = -11 x 10^-6 m

R = 8.31 J mol^-1 K^-1

Density of water, ρ' = 1000 kg m^-3

Using the relation

We have,

⇒ 2.333 = exp (1.241 x 10^-24 N_A)

Taking natural log of both sides, we get

log 2.333 = 1.241 x 10^-24 N_A

or,

The correct value of N_A is 6.02 x 10²³ mol^-1.