Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL–1?

Answer :-

Step-1  find the given values

Mass % of nitric acid  = 68%

Density of nitric acid  = 1.504 g mL-1

Step -2

Let take total mass of solutions = 100    (why ?)

In this problem mass % is given whenever % is given take total quantity as 100 g ,

So  nitric acid =  68% of 100 g   = 68 g

Water will = 100 – 68  =  32 g

Step-3

We have to find molarity

Molarity(m) =  number of moles of solute / volume of solution in liter           ………..(1)

Here we have to find number of moles of solute  and volume of solution in liter                    

 Number of moles of solute = mass of solute /molar mass of solute             …………(2)

So molar mass of nitric acid HNO3  = 1*H + 1*N + 3*O  = 1*1 + 1*14 + 3*16  = 1+ 14 + 48 = 63 g

Plug the value in equation (2)  we get

Number of moles  = 68/63 = 1.079 moles

Volume = mass / density

              = 100g/1.504 g mL–1    =  66.5 ml

Now convert it in liter we get 

                                    66.5 ml * 1 liter / 1000ml =  0.0665 mL

Plug the value in equation first now we get

Molarity  = 1.079/ 0.0665  =16.23 M