What is the $"molality"$ of pure water?

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Answer 1 · 2017-05-08T04:06:12

Large........... $>50*mol*kg^-1$

$"Molality"="Moles of solute"/"Kilograms of solvent"$ ,

and so if we assume a $1*kg$ mass of water,............(what is the volume?).........then

$"Molality"=((1000*g)/(18.01*g*mol^-1))/(1*kg)=??mol*kg^-1.$

And this value is quite comparable to water's $"molarity"$ ?

What is the "molality""molality" of pure water?