Question

(A) Calculate the molarity of commercially available concentrated nitric acid reagent given the following data:

Molecular formula: HNO3

Molar mass = 63.01 g/mol

Purity/Reagent Concentrate: 70.9% (w/w)

Specific gravity @20 degrees Celsius = 1.423

(Hint: Using the specific gravity, calculate the mass of one liter solution. Of this mass, only 70.9% is pure HNO3; the remainder being water).

(B) To make 212 mL of 0.15 M solution of nitric acid, what volume of the concentrated nitric acid will be needed?

Final answer: (A) Molarity of concentrated HNO3 = _ M. (B) Volume of conconcentrated HNO3 required = _ m

Answer 1

(A)

suppose we take one litre of reagent

then mass of reagent = ( volume of reagent) (density of reagent)

specific gravity = (density of reagent) / (density of water)

density of reagent = (specific gravity) (density of water)

   density of reagent = 1.423 1000 (kg/m³)

= 1423 kg/m³

mass of reagent = ( volume of reagent) (density of reagent)

mass of reagent = 1(L) 10^-3 (m³/L) 1423 (kg/m³) = 1.423 kg or 1423 gm

mass of nitric acid in the reagent =

moles of nitric acid = mass (gm) / molar mass

    

= 16.01 mol

molarity =

  

molarity = 16.01 M

(B) we need to make 212 mL of 0.15 M nitric acid solution.

moles of nitric acid = molarity volume of solution (litre)

= 0.15 (M) 212 (mL) 10^-3 (L/mL)

= 0.0318 mol of nitric acid.

the concentrated nitric acid reagent have a molarity of 16.01 M ( calculated in part (A)).

so to get 0.0318 mol of nitric acid

moles of nitric acid = molarity volume of solution (Litre)

0.0318 (mol) = 16.01 (M) volume of solution (Litre)

   volume of concentrated nitric acid required = 1.986 10^-3 L

or = 1.986 mL