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^-1# solution of ethanolamine, #H_2NCH_2CH_2OH#?