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

1 Answer
Nov 14, 2015

Answer:

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)#.