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The alternate value for C'_{S} is obtained from a fit to measured data found here. This is a fit to measured data that covers much higher resistivities. The equation is log(C'_{s})=(log(1553)log(ρ))/0.7465, where C'_{S} is in g/l and ρ is in Ohmcm.  
You should use deionized water and copper electrodes to reduce resistance change over time.  
Example: Suppose you want to make a 1000 ohm resistor out of 1" innerdiameter Tygon tubing with copper electrodes inserted at each end with 1 foot between electrodes. The calculator gives C'_{s}=19.653 g/l and C'_{w}= 991.41 g/l. So, to make one liter of solution, fill a jar with 991.41 g of water and 19.653 g of CuSO_{4}*5H_{2}O (or make a smaller batch keeping the same proportions). Note: you could just add the amount, C'_{s}, to one liter of water with negligible error (you have to do this for the alternate version of C'_{s}). Also, if you want to make just enough then use the volume (in liters) from the calculator and multiply this by C'_{s} to get the amount of CuSO_{4}*5H_{2}O you need, then fill with water.  
CuSO_{4}*5H_{2}O is know by several names (copper sulfate, cupric sulfate, cupric sulfate pentahydrate,...) and appears as blue crystals.  
Calculator assumes solution temperature of 20^{o}C (68^{o}F). Actual resistance is a function of water temperature.  
In the table below, the left three columns are data from the CRC Handbook of Chemistry and Physics. H% is the percent of hydrate (CuSO_{4}*5H_{2}O) by weight in the solution. D^{20}_{4} is the relative density of the solution at 20^{o}C compared to water at 4^{o}C. γ is the specific conductance at 20^{o}C. The right three columns are derived from the others for this calculator.  
 
Note that CRC Handbook gives values for C_{s} and C_{w}, but these count the 5H_{2}O as part of the solution, not the solute, which would result in a slight error if not accounted for (this is why my values have the prime symbol).  
Density of 20^{o}C water is actually 0.99823 [g/ml] from this table.  
If you want to use anhydrous CuSO4 (gray powder) then multiply amount of
solute by 0.639 (you need less) and use more water given by the difference.

Copyright ©2003 Raymond J. Allen