Help with Drift Velocity of Ions Isaac Physics Question

Q: In an experiment, a current of 3.5A is being passed through a copper sulphate solution in a 10cm cubical container, with the electrical terminals being opposite faces. This contains equal numbers of Cu 2+ and SO4 2− ions which have respectively +2 and −2 electron charge units. Assuming that the two ions have equal speed in the solution, and that there are 6.0×10^26 of each per cubic metre of the solution, work out their mean speed to 2 significant figures.

My Working So Far:
10x10=100cm^2
100cm^2=0.01m^2

10x10x10=1000cm^3
1000cm^3=1x10^-3m^3
1x10^-3x6x10^26=6x10^23

3.5/(6x10^23)x(2x1.6x10^-19)x(0.01)=1.8x10^-3 m/s

This is incorrect and I am stuck at this point and do not know where I went wrong. Thanks for any help provided.

Well you're trying to do the right calculation
but there are a couple of things

1. the n in I=nvqA is the number density... the number of charge carriers per cubic meter - you don't want to scale this number down to the number in the sample on test - it's the same for any amount of the same solution.

2. each of the two types of ion is carrying a charge of size 2*1.6E-19 and there are 6.0E26 of EACH ion - so you're also out by a factor of 2.

Q: In an experiment, a current of 3.5A is being passed through a copper sulphate solution in a 10cm cubical container, with the electrical terminals being opposite faces. This contains equal numbers of Cu 2+ and SO4 2− ions which have respectively +2 and −2 electron charge units. Assuming that the two ions have equal speed in the solution, and that there are 6.0×10^26 of each per cubic metre of the solution, work out their mean speed to 2 significant figures.

My Working So Far:
10x10=100cm^2
100cm^2=0.01m^2

10x10x10=1000cm^3
1000cm^3=1x10^-3m^3
1x10^-3x6x10^26=6x10^23

3.5/(6x10^23)x(2x1.6x10^-19)x(0.01)=1.8x10^-3 m/s

This is incorrect and I am stuck at this point and do not know where I went wrong. Thanks for any help provided.

I = nQva

So
n= Charge carrier desnity (6x10^24)
Q = Charge.
A = cross sectional area so = .01 m^2

nxq = the total charge of all the charge carriers. We have 4 charge carriers so this will be 4x 1.6 x10^-19
Rearange the formulae to v=i/nQa and subsititure the values

comes out as stated physics student 31 stated.

how do you get the charge carrier density?