What is the cost of 1.00 g of calcium ions as provided by this brand of dry milk? A student determined the $\ce{Ca^{2+}}$ ion content in a sample of nonfat dry milk. For complete titration of the $\ce{Ca^{2+}}$ ion present, a sample weighting $1.483 ~\mathrm{g}$ required $41.33 ~\mathrm{mL}$ of $1.183 \times 10^{-2} ~\mathrm{M}$ EDTA solution. Calculate the mass of the $\ce{Ca^{2+}}$ ions in the titrated sample.


So I did:

$41.33 ~\mathrm{mL} = 0.04133 \mathrm{L}$

$(1.183 \times 10^{-2}~\mathrm{mol\, L^{-1}})\times (0.04133 \mathrm{L}) = 4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}$

$(4.89 \times 10^{-4} ~\mathrm{mol}\, \ce{Ca^{2+}}) \times 40.078 ~\mathrm{g\, (mol\, \ce{Ca^{2+}})^{-1}} = 0.0196 ~\mathrm{g}\, \ce{Ca^{2+}}$

My question is:

A $25.06$ oz box of the powdered milk analyzed above costs $5.46$. 


  Calculate the cost of $\pu{1.00 g}$ of $\ce{Ca^{2+}}$ ion as provided by this brand of dry milk.


How do I do that?
First you calculate the value $v$ of the powder per gram:
$$ 25.06~\mathrm{oz} = 710.44~\mathrm{g} \; \hat{=} \; 5.46~\$ \Rightarrow v=0.007685~\mathrm{\$\, (g\, Powder)^{-1}}$$

Then you calculate the mass of powder ($m$) that contains one gram of calcium ions:
$$ m=\frac{1.483~\mathrm{g\, Powder}}{0.0196~\mathrm{g\, \ce{Ca^{2+}}}} = 75.66~\mathrm{\frac{g\, Powder}{g\, \ce{Ca^{2+}}}} $$

Finally, you multiply the two results for the price $p$:
$$ p = v\times m = 0.58~\mathrm{\$\, (g\, \ce{Ca^{2+}})^{-1}}$$