Question

concentrated nitric acid used in laboratory work is 68%

5 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/mol?

Answer 1

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. From statement, its clear that 68 g of nitric acid is mixed in 100 g of the solution.

Given, density  = 1.504g/mL-1    

Density = mass / volume

1.504g/mL-1 = 1000 / volume

Volume = 1000/1.504 = 66.49 mL

From the formula of nitric acid (HNO3), the molar mass would be = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol - 1

Therefore, We can calculate number of moles of HNO3 = 68 / 63 mol  

= 1.08 mol

Molarity of sample = (Number of moles / Volume) x mass

                           

                               = (1.08/66.49) x 1000 = 16.24 M

The above answer is correct if you don't use numeric 5 in the question otherwise question is incorrect according to me.

Answer 2

How the mass come 1000 here?