Question

A sample of concentrated nitric acid is 69% by mass and has a density of 1.41 gm/ml.
Calculate the molarity of this sample. Given: at. Mass of H=1, O=16 & N=14.​

Answer 1

Answer:

Given that Mass percent of nitric acid in the sample = 69 %

When mass % is given always take total mass = 100 g

Mass of nitric acid = 69% of 100 g = 69 g

Molar mass of nitric acid (HNO3)

= {1 + 14 + 3(16)} g mol-1 = 1 + 14 + 48 = 63 g mol-1

Use formula density = mass / volume

Given that density = 1.41 g ml-1

Total mass = 100 g

Plug the values in above formula

Volume of nitric acid = 100/1.41 = 70.92 ml

Concentration in moles per litre = number of moles / volume in liter …(1)

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

Molar mass of nitric acid (HNO3) = 1×1 + 14 + 3 × 16 = 63 amu

Plug the value in equation (2) we get

Number of moles of HNO3 = 69/63 = 1.095

Volume in liter = 70.92 ml = 0.07092 liter (divide by 1000 to convert in liter)

Plug the value in equation (1), we get

Concentration in moles per liter = 1.095/ 0.07092 = 15.44 mol / liter