Question

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

Answer 1

Approx. `1.8` `mol*L^-1` with respect to `NaI`.

Explanation:

`"Molarity"` `=` `"Moles of solute"/"Volume of solution"`

`=` `(107*g)/(149.9*g*mol^-1)xx1/(0.250*L)` `=` `??*mol*L^-1`.

Note that we would not expect the volume of solution to change markedly from the volume of the solvent used.