Question

What is the molarity of .50 g of `Na` dissolved in a 1.5 L solution?

Answer 1

In aqueous solution, approx. `0.015*mol*L^-1` with respect to sodium ion.

Explanation:

`Na+H_2O rarr NaOH(aq) + 1/2H_2(g)uarr`

`"Moles of sodium metal"` `=` `(0.50*g)/(22.99*g*mol^-1)`

`"Concentration"` `=` `"Amount in moles"/"Volume of solution (L)"`

`=` `(0.50*g)/(22.99*g*mol^-1)xx1/(1.5*L)`

`=` `??mol*L^-1`.

Under standard conditions (`"298K, 1 atm"`), what volume of gas will be evolved?

Note that I could have dissolved the sodium in mercury metal; in fact this is often done in synthetic laboratories. Now that would be quite a heavy solution.