Question

What is the molarity of 0.080 moles `NaHCO_3` in 1250 mL of solution?

Answer 1

`"Molarity = 0.064"*mol*L^-1`

Explanation:

`"Molarity"` `=` `"Amount of substance (moles)"/"Volume of solution (L)"`

`=` `(0.080*mol)/(1.250*L)` `=` `0.064"*mol*L^-1` with respect to sodium bicarbonate.