What is the solubility of #KCl# at 25°C?
1 Answer
Aug 4, 2017
Once you look it up, you should be telling me that... I get
#ul(s_("KCl")) = (360 cancel"g KCl" xx "1 mol"/(74.5513 cancel"g KCl"))/(cancel"1 kg water" xx (1000 cancel"g")/cancel"1 kg" xx "1 L"/(997.0749 cancel"g")#
#=# #ul"4.815 M"#
Or maybe...
#(360 cancel"g KCl")/"kg water" xx "1000 mg"/cancel"1 g" = ul(3.60 xx 10^5 " ppm")#
Or perhaps, molality?
#(360 cancel"g KCl" xx "1 mol"/(74.5513 cancel"g KCl"))/("1 kg water") = ul("4.829 mol/kg")# , or#"molal"# , or#"m"#
How about... percent by mass ("weight percent")?
#"360 g KCl"/cancel"kg water" xx cancel"1 kg"/"1000 g" xx 100% = ul(36% "w/w")#