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.

Related questions
See all questions in Solvation and Dissociation
Impact of this question
1854 views around the world
You can reuse this answer
Creative Commons License