Question

You can purchase nitric acid in a concentrated form that is 70.3% HNO(3) by mass and has a density of 1.41 g/mL. Describe exactly how to prepare 1.15 L of 0.100 M HNO(3) from the concentrated solution.

Please explain thoroughly.

Answer 1

Moles of HNO₃ present in 1.15 L of 0.10 M solution

                  moles = molarity* volume in liters

                            = 0.10 M* 1.15 L

                            = 0.115 mol

Mass of HNO₃ = moles* molar mass

                         = 0.115 mol *63 g/mol

                         = 7.245 g

Mass of 70.3 % HNO₃ solution which contains 7.245 g HNO₃

                    = 7.245 g HNO₃ * (100g Solution/70.3 g HNO₃)

                    = 10.306 g

Volume of concentrated solution required = mass/density

                                                                = 10.306 g / 1.41 g/mL

                                                                = 7.3 mL

Total volume of final solution = 1.15 L

                                              = 1150 mL

   Volume of water required to prepare final solution = 1150 mL - 7.3 mL

                                                                               =1142.7 mL

    Pocedure: If we mix 7.3 mL of 70.3 % HNO₃ solution and 1142.7 mL distilled water then   

    we will get 1150 mL(1.15L) of 0.10 M HNO₃ solution.