How do we make a $1molL^-1$ solution of ethanolamine, $H_2NCH_2CH_2OH$ ?

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Answer 1 · 2017-04-18T17:30:52

$"Molarity"$ $=$ $"Moles of solute"/"Volume of solution"$

And so we take a $61.08*g$ mass of ethanolamine, and make it up to a $1*L$ volume in a volumetric flask:

$"Molarity"=((61.08*g)/(61.08*g*mol^-1))/(1.00*L)=1.00*mol*L^-1$

Would this give rise to a basic or acidic solution?

How do we make a 1*mol*L^-11*mol*L^-1 solution of ethanolamine, H_2NCH_2CH_2OHH_2NCH_2CH_2OH?