Question

9. [1pt] When current flows in an ionic solution, both negative and positive ions are charge carriers. In the dilute limit, the resistivity of the solution is inversely proportional to the concentration. For example, the resistivity of salt water solution at 25 °C is ρ = 8.0645/[NaCl] Ω·cm·mol/L, where [NaCl] is the concentration of salt in the water, in moles per litre.

a)Calculate the resistance of a cylinder of salt water (in a plastic tube) with radius r = 1.70 cm, length L = 13.90 cm, and [NaCl] = 0.10 mol/L.

b)In an in-class demonstration, a pickle was “electrocuted”. Assuming the pickle to be equivalent to a cylinder of water with resistance, R, as calculated above, what current, I, would flow through the pickle for an applied potential difference of V = 64 V (alternating current)?

c)For the situation described, with a current I flowing through the water, what is the drift speed of Na⁺ ions while the current flows?

Answer 1

a.

Correct ρ = 8.0645/[NaCl] Ω·cm·mol/L to ρ = 8.0645 Ω·cm·mol/L / [NaCl]

The resistance R = ρ · L / (πr²)

    R = (8.0645 Ω·cm·mol/L / 0.10 mol/L) x 13.90 cm / (3.14x1.70² cm² ) =

        = 1.532 Ω

b. I =U/R = 64V/1.53 Ω = 41.8 A = 42 A

c.By definition 1A = 1C/s .

Here 42 A = 42 C/s

1 mol of electric charge = 96500 C

1 mol NaCl has 1 mol Na+ and 1 mol Cl- .

In 1 cm³ sol NaCl 0.1 M there are 0.1 mol/L /1000cm³/L = 1x10^-4 mol/cm³ Na⁺ equivalent to a charge of

   96500 C/mol x 1x10^-4 mol/cm³ = 9.65 C/cm³ (charge concentration)

I = charge/time = charge concentration x area x drift speed

42 C/s = 9.6 C/cm³ x (3.14x1.70² cm² ) x v

The drift speed v = 0.48 cm/s