# Question d42c0

May 2, 2017

A hardness of 320.7 ppm corresponds to 0.3556 g of ${\text{CaCl}}_{2}$ per litre.

#### Explanation:

A hardness of 1 ppm refers to the concentration of $\text{Ca"^"2+}$ expressed as 1 ppm of ${\text{CaCO}}_{3}$.

$\text{1 ppm of hardness" = "1 g"/(10^6color(white)(l) "g") = "1 mg"/(10^3color(white)(l) "g") = "1 mg/L}$

$\text{320.7 ppm of hardness" = "320.7 mg/L" = "0.3207 g/L}$

Thus, you need 0.3207 g of $\text{CaCO"_3}$ per litre.

For ${\text{CaCl}}_{2} , {M}_{\textrm{r}} = 110.98$.

For ${\text{CaCO}}_{3} , {M}_{\textrm{r}} = 100.09$.

Thus, to get the same amount of $\text{Ca"^"2+}$, you need

0.3207 color(red)(cancel(color(black)("g CaCO"_3))) × "110.98 g CaCl"_2/(100.09 color(red)(cancel(color(black)("g CaCO"_3)))) = "0.3556 g CaCl"_2#