$**$PLEASE DO ALL AND NOT THE FIRST ONE I COMPLETED..."$**$

In the titration of $25\mathrm{.}00mL$ of a water sample, it took $19\mathrm{.}490mL$ of $3\mathrm{.}780{10}^{-3}M$ EDTA solution to reach the endpoint.

Calculate the number of moles of EDTA required to titrate the water sample. $($enter your answer with $3$ significant figures$)$

$7\mathrm{.}37{10}^{-5}$

You are correct.

Your receipt no$.$ is $153-1086(?)$ Previous Tries ${?}_{(})$

The total hardness is due to one or a combination of $C{a}^{2+},M{g}^{2+},$ and $F{e}^{2+}$ in your sample. It is convenient to express this hardness as though it was entirely due to $C{a}^{2+}.$ Making this assumption, determine the number of mole of $C{a}^{2+}$ present in the bottled water sample titrated. $($enter your answer with $3$ significant figures$)$

Tries $1/99$ Previous Tries

The total hardness is always listed in parts$-$per$-$million $($ppm$)$ of $CaC{O}_3($or mgCaC$\frac{O_3}{K}g{H}_{2}O).$ Since the density $0$ water is $1\mathrm{.}0\frac{g}{m}L,$ one ppm would be the same as the number of $mg$ of $CaC{O}_3$ per liter of water. Determine the number of moles of $CaC{O}_3$ present in the titrated sample of water, assuming that all the $C{a}^{2+}$ combines with $C{O}_3^{2-}.($enter your answer with $3$ significant figures$)$

Tries $\frac{0}{99}$

Calculate the number of grams of $CaC{O}_3$ present and convert to $mg.($enter your answer with $3$ significant figures$)$

Tries $0/99$

Convert the number of $mL$ of bottled water used in each sample titration to Liters. $($enter your answer with $3$ significant figures$)$

Tries $0/99$

Calculate the ppm of $CaC{O}_3=$mgCaCO $/$ Liters $H_{2}O$ used. $($enter your answer with $3$ significant figures$)$