Question

The molecular weight of Na is 58.44 grams/mole. To prepare a 0.50 M solution: how many grams of NaCI would you need in 1,000 mL of water?

Answer 1

`"Molarity"` `=` `"Moles of solute"/"Volume of solution"`, so here we need `29.22*g` of `NaCl`

Explanation:

And thus, `(29.22*g)/(58.44*g*mol^-1)xx1/(1*L)` `=` `0.500*mol*L^-1`.

You might ask as to whether the volume of water would change (up or down upon dissolution of the salt. The volume change would be imperceptible.