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

Apr 13, 2017

Well, $\text{lithium nitrate}$ is the solute.........

#### Explanation:

And we can quote the molar concentration with respect to $L i N {O}_{3}$. And we can calculate the concentration of the solution with respect to this solute...........

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

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

$= 1.54 \cdot m o l \cdot {L}^{-} 1$ because $\frac{1}{\frac{1}{x}} = x$.

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