Question

What is the molarity of a solution prepared by dissolving 0.450 of `HNO_3` in 300 mL of water?

Answer 1

Moles of nitric acid: `(0.450*g)/(63.01*g*mol^(-1))`. This is dissolved in `300` `mL` of water. I have assumed that you mean `0.450* g` of nitric acid.

Explanation:

`"Molarity"` `=` `("moles of solute")/("volume of solution")`.

`((0.450*g)/(63.01*g*mol^(-1)))/(0.300*L)` `=` `??` `mol*L^(-1)`

Please do the math, and report the answer in `mol*L^(-1)`.