Hardness of water using EDTA calculation issue

I have been asked to work out the hardness of water in German degrees and this is what I did:


I got a sample of water I didn’t measure how much I put into my volumetric flask perhaps it was $\pu{8.5ml}$ but I didn’t think I would need it, but I do use it which made me believe I was wrong. 


The sample was made up to $\pu{100ml}$. $\pu{10ml}$ of this dilution was used as the analyte along with buffer and the indicator.


I Titrated $\pu{9.07ml}$ of EDTA with a concentration of $\pu{0.0185mol/dm^3}$


So the calculation I did was:


$$\begin{align} n(\ce{edta})&= (9.07\times0.0185)/1000= 0.000167795\\
n(\ce{Ca^2+}) &= n(\ce{edta})\end{align}$$
So then I find the mass of Ca:
$$m(\ce{Ca^2+}) = 40\times0.000167795= \pu{0.0067188g}$$


Now my issue:


So $\pu{0.0067188g}$ of Ca in the $\pu{10ml}$ therefore in the $\pu{100ml}$ there will be $10$ times that, as it was a dilution. 
So $\pu{0.067188g}$ which is $\pu{67.118mg}$.


So there is $\pu{67.118mg}$ of Ca in my $\pu{8.5ml}$ sample.  

Now a $1^\circ\ \mathrm{dH}$ is $\pu{10mg}$ of Ca per $\pu{1000ml}$.
So is there $\pu{7896.2mg}$ in $\pu{1000ml}$ of sample — 
therefore having $789.6^\circ\ \mathrm{dH}$ of hardness? 




Please always, always, always, always use units. An amount is not equal to a simple number like in your first calculation. Likewise, the multiplication of two simple numbers does not give a value in grams like in your second calculation.


That said, as far as I can tell your calculations add up. (It took me a few tries since you juggle with units so liberally — another reason to use them *consistently.)* So maybe you were not given tap water, maybe you grossly misestimated the volume you added to the flask initially or maybe your titration was wrong. What exactly happened is nothing we can answer.


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