Ion solution drift speed? (1 Viewer)

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1. The problem statement, all variables and given/known data

A. 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. Calculate the resistance of a cylinder of salt water (in a plastic tube) with radius r = 1.30 cm, length L = 13.80 cm, and [NaCl] = 0.29 mol/L.

B. If a potential difference of V = 69 V (alternating current) is applied, what will the current be?

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?

2. Relevant equations
Area of a circle= pi*r2
I=n*q*vd*A, where n is the number of charge carriers per unit volume, q is the charge on the charge carriers, vd is the drift velocity, and A is the cross-sectional area.

3. The attempt at a solution

For A. Simply using the formula, I get 72.3 Ohms

For B. Again, via formula use, I get .955 Amperes.

For C. I have no idea what to use for n. I would be .955, q I presume would be 1.602*10-19, A would be the pi*(1.30/100)2 (since I presume my units would require the area to be given in m^2?)

I tried using the given molarity of 0.29 mol/L=0.29mol/m^3, and multplying that by Avegadros number (6.02*1023) to get the number of ions per unit volume, but that didn't seem to give me the correct answer.

Anything I'm missing here, or perhaps theres another formula I should be using to get at vd?
There are 2 moles of ions (Na+ & Cl-) per moles of NaCl. I hope you can assume their velocities to be the same.
Since it hasn't said otherwise, I would presume it's a fair assumption if theres no other clear way to get the answer.

The fact theres two moles in each case however, would mean what in terms of the equation? Would it double both the value of n and q, since theres twice as many charge carriers, and the total of the absolute charges is doubled? Or would it only double n?

Or perhaps instead, I should be halving 'I', since I only care about the current resultant from the Na+ ions, instead of the total current coming from both parts?

All the examples I've seen only discuss electron flow densisties rather than actual ions, so I'm unsusre of how a change such as this should influence the formula.
Got it. Double n to account for the fact theres two ions. Leave q alone.

And more importantly, 1L=0.001m3, not 1L=1m3 as I'd initially been treating it.

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