When 21.2 g of #LiNO_3# is dissolved to make 200.0 mL of solution, what is #LiNO_3#?

1 Answer
anor277
Apr 13, 2017

Well, #"lithium nitrate"# is the solute.........

Explanation:

And we can quote the molar concentration with respect to #LiNO_3#. And we can calculate the concentration of the solution with respect to this solute...........

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#

#=((21.2*cancelg)/(68.95*cancelg*mol^-1))/(200*cancel"mL"xx10^-3*L*cancel"mL"^-1)=1.54*(1/(1/"mol"))*L^-1#.

#=1.54*mol*L^-1# because #1/(1/x)=x#.

Are my calculations right? All care taken but no responsibility admitted.

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