Question

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

Answer 1

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.