Question #37f4a

1 Answer
Truong-Son N.
Feb 13, 2018

Well, you didn't mention the temperature, so we'll have to assume what density water has. At #25^@ "C"#, #rho_(H_2O(l)) = "0.9970749 g/mL"#, or #"kg/L"#.

The unit of #"ppm"#, parts per million, could be written

#"mg solute"/"kg solution"#, #(mu"g solute")/("g solution")#, etc.

Therefore,

#"100 mg"/(2 cancel("L H"_2"O") xx "0.9970749 kg"/cancel"L" + "100 mg solute") = "100 mg"/"1.9941498 + 0.0001 kg solution"#

#~~ "100 mg solute"/("1.994 kg solution")#

#=# #color(blue)("50 ppm")#

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