Question

Question #d42c0

Answer 1

A hardness of 320.7 ppm corresponds to 0.3556 g of `"CaCl"_2` per litre.

Explanation:

A hardness of 1 ppm refers to the concentration of `"Ca"^"2+"` expressed as 1 ppm of `"CaCO"_3`.

`"1 ppm of hardness" = "1 g"/(10^6color(white)(l) "g") = "1 mg"/(10^3color(white)(l) "g") = "1 mg/L"`

∴ `"320.7 ppm of hardness" = "320.7 mg/L" = "0.3207 g/L"`

Thus, you need 0.3207 g of `"CaCO"_3"` per litre.

For `"CaCl"_2, M_text(r) = 110.98`.

For `"CaCO"_3, M_text(r) = 100.09`.

Thus, to get the same amount of `"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`