# What is the molarity of a solution prepared by dissolving 107 g NaI in 0.250 L?

Approx. $1.8$ $m o l \cdot {L}^{-} 1$ with respect to $N a I$.
$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$
$=$ $\frac{107 \cdot g}{149.9 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.250 \cdot L}$ $=$ ??*mol*L^-1.