***Only requirement is bottom chart filled out with concentrations of unknown substance with the below given conductivities***Ocean water was 20 ml and diluted with 80 ml of distilled water #Unknown SampleConductivity (µS/cm)1.Distilled Water42.Tap Water6373.Outside Faucet or Hose Water6504.Bottled Water345.Rain Water376.Ocean Water8817All experimental data comes in as individual discreet data points with gaps between those data points.Its highly unlikely the conductivity of an unknown will land on exactly, or match to, the conductivities of the above calibration conductivities. For that reason, its important to analyze the data properly so as to interpolate the unknown measurements with the above calibration curve to properly and accurately translate the unknown conductivity to its unknown concentration, then becoming known within proper error, precision, and significant figures. The most common method is to fit the pattern of the graphed and individual discreet data points to a known mathematical expression, like a line, polynomial, or function, like sine, exponential, or logarithm. Fitting discreet data to a mathematical function connects the dots and fills the gaps between the data points.It can be proven and derived that conductivity data follows a polynomial of the form:L=aC32+bC+cwhere L is the conductivity, C is the concentration of the solution, and a, b, and c are fitting parameters that trace back to physically relavent constants based on the charges and mobilities of the ions within the aqueous solution. There is no mathematically exact or analytical solution for a polynomial to the three halves (3/2) order.MS Excel, with proper training, can fit and numerically solve such a polynomial, but in this class we will back track to a simpler and old-school technique.The old-school technique is to grab the two calibration points on either side of the unknown measurement. In other words, grab the two points from the calibration data that are between the unknown conductivity.Use those two points and solve the line from those two points. So (C1, L1) and (C2, L2) convert to the equation of a line with L = mC + b (y = mx + b) where using the two points you’ll derive or find the slope, m, and the y-intercept, b. You should recognize or refamiliarize yourself with the equation of a line from your old math classes.Briefly, find the slope, m, by:m=riserun=ΔyΔx=ΔLΔC=L2−L1C2−C1where Δ means the “change in” and is always “the final minus the initial”. And the y-intercept is found by:b=y1−mx1=L1−mC1where m is the latter derived slope. Once m and b are numerically found then you’ll have your equation of a line, or:y=mx+borL=mC+bwhere once m and b are found and just numbers, then once you plug in the unknown conductivity, L, you can solve and find the unknown concentration, C, which is now known, accurately, precisely, with specific significant figures, and within error limits. An example of this line fitting technique between two data points is laid out under the Data and Observations section of the Formal Laboratory Report, an Example of this lab manual.From the latter technique, analyze your unknown sample conductivities and report your derived concentrations below.Remember, if you had to dilute your unknown sample to get a measureable reading from your probe which derived a concentration as above, then be sure to multiple that concentration by the dilution factor to bring it up to the final concentration of the undiluted sample. Report that final undiluted concentration below.#Unknown SampleConcentration (M)1.Distilled Water2.Tap Water3.Outside Faucet or Hose Water4.Bottled Water5.Rain Water6.Ocean Water

Concentration of Unknown Substance with Given Conductivity Worksheet