Question

If 25.0 g of `K_2CO_3`, potassium carbonate, are dissolved in `450 cm^3` of solution, what is the molarity?

Answer 1

Approx. `0.4*mol*L^-1`

Explanation:

`"Molarity"` `=` `"Moles of potassium carbonate (mol)"/"Volume of solution (L)"`

`=((25*g)/(138.21*g*mol^-1))/(0.450*L)~=0.4*mol*L^-1`

Note that when we do the calculation this wa we automatically get appropriate units. We seek the molarity, which should properly have units of `mol*L^-1`, and lo and behold, the expression gives us such units as the grams cancel out, and the `mol^-1` is inverted to give `1/(1/(mol))=mol`.......