# Density , Relative Density , Density of Mixture

Density , Relative Density of Substance:
Matter exists in three states solid, liquid and gas. Any state of matter that can flow is a fluid. Liquids and gases are therefore, referred to as fluids.

### What is Density ?

Density of a substance is the mass per unit volume . Homogenous bodies have uniform density .

$\large Density = \frac{Mass}{Volume}$

The unit of density is kg/m3 in S.I. system. Dimension of density is given by [ρ] = = [M] [L-3].

### What is the Relation between Density & Relative Density or Specific Gravity ?

Relative density of a substance is the ratio of the density of the substance to the density of pure water (ρw = 1000 kg/m3) at 4°C.

$\displaystyle \rho_r = \frac{\rho}{\rho_w}$

“Relative density is dimensionless ”

Solved Example : A hollow metallic sphere has inner and outer radii respectively as 5 cm and 10 cm. If the mass of the sphere is 2.5 kg, find (a) density of the material (b) relative density of the material of the sphere.

Solution: The volume of the material of the sphere is

$\displaystyle V = \frac{4}{3}\pi( r_2^3 – r_1^3)$

$\displaystyle V = \frac{4}{3}\times 3.14( (0.1)^3 – (0.05)^3)$

= 0.00367 m3

(a) Therefore, density of the material of the sphere is

$\displaystyle \rho = \frac{M}{V}$

$\displaystyle \rho = \frac{2.5}{0.00367}$

= 681.2 kg/m3

(b) $\displaystyle \rho_r = \frac{\rho}{\rho_w}$

$\displaystyle \rho_r = \frac{681.2}{1000}$

= 0.6812

## How to Calculate Density of a mixture ?

Let a number of substances of masses M1, M2, M3 etc., and densities ρ1, ρ2, ρ3, etc. respectively be mixed together.

The total mass of the mixture is = M1 + M2 + M3 + ….

The total volume $\displaystyle = \frac{M_1}{\rho_1} + \frac{M_2}{\rho_2} + \frac{M_3}{\rho_3} + …..$ ;
provided that the substances retain their individual states within the mixture.

Therefore, the density of the mixture is

$\displaystyle \rho_{mix} = \frac{M_1 + M_2 + M_3 + ….}{\frac{M_1}{\rho_1} + \frac{M_2}{\rho_2} + \frac{M_3}{\rho_3} +….}$

#### The same expression may written in terms of the volumes:

The density of the mixture is

$\displaystyle \rho_{mix} = \frac{\rho_1 V_1 + \rho_2 V_2 + ….}{V_1 + V_2 + ….}$

where V1, V2, V3 … represent the volumes of substances of densities ρ1, ρ2, ρ3 . . . in the mixture

Solved Example : Two liquids of densities 2.5 gm/cm3 and 0.8 gm/cm3 are taken in the ratio of their masses as 2:3 respectively. Find the average density of the liquid combination.

Solution: Let the masses be ‘ 2 m ‘ gm and ‘ 3 m ‘ gm respectively.

Therefore, the volume of the first liquid of density 2.5 gm/cm3 is V1 = 2m/2.5 cm3

That of the second liquid is V2 = 3m/0.8 cm3

Total volume $\displaystyle V = V_1 + V_2$

$\displaystyle V = \frac{2m}{2.5} + \frac{3m}{0.8}$

Total mass = 2m + 3m = 5 m gm

Therefore, the average density

$\displaystyle \rho_{av} = \frac{5m}{V}$

$\displaystyle \rho_{av} = \frac{5m}{\frac{2m}{2.5} + \frac{3m}{0.8}}$

$\displaystyle \rho_{av} = \frac{10}{9.1}$

= 1.09 gm/cm3

Solved Example: Two miscible liquids of densities 1.2 gm/cc and 1.4 gm/cc are mixed with a proportion ratio of their volumes equal to 3:5. What is the density of resulting liquid?

Solution: density of resulting liquid is

$\large \rho = \frac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2}$

$\large \rho = \frac{\rho_1 \frac{V_1}{V_2} + \rho_2 }{\frac{V_1}{V_2}+ 1}$

$\large \rho = \frac{1.2 \times \frac{3}{5} + 1.4 }{\frac{3}{5}+ 1}$

$\large \rho = \frac{3.6 + 7}{8} = \frac{10.6}{8}$

ρ ≅ 1.32