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Answer 1 · 2017-05-02T18:48:41

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

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#