What is the solubility of #KCl# at 25°C?

Answer 1

Once you look it up, you should be telling me that... I get #"360 g KCl"/"kg water"# at #25^@ "C"# off of Wikipedia. And hence, the solubility in molarities is...

#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... #"ppm"#?

#(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")#

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Answer 2

Peyton Adams

The solubility of KCl at 25°C is approximately 34 grams per 100 grams of water.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.