Question

Three liquids of densities d,2d and 3d are mixed in equal proportions of weights. The relative density of the mixture is

Answer 1

Answer: The relative density of mixture is $\frac{18d}{11}$

Explanation:

The equation used to calculate density of a substance is given by:

$\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}$

We are given:

Density of liquid 1 = d

Density of liquid 2 = 2d

Density of liquid 3 = 3d

To calculate the relative density of a mixture, we use the equation:

$\rho_{mix}=\frac{m_1+m_2+m_3}{V_1+V_2+V_3}$

where,

$m_1\text{ and }V_1$ are the mass and volume of liquid 1

$m_2\text{ and }V_2$ are the mass and volume of liquid 2

$m_3\text{ and }V_3$ are the mass and volume of liquid 3

We are given:

Mass of all the three liquid are equal to 'm' because they are mixed in equal proportions of weight.

$m_1=m\\V_1=\frac{m}{d}\\\\m_2=m\\V_2=\frac{m}{2d}\\\\m_3=m\\V_3=\frac{m}{3d}$

Putting values in above equation, we get:

$\rho_{mix}=\frac{m+m+m}{\frac{m}{d}+\frac{m}{2d}+\frac{m}{3d}}\\\\\rho_{mix}=\frac{3\times 6d}{11}=\frac{18d}{11}$

Hence, the relative density of mixture is $\frac{18d}{11}$

Answer 2

Answer:

substance}}Density of a substance=

Volume of a substance

Mass of a substance

We are given:

Density of liquid 1 = d

Density of liquid 2 = 2d

Density of liquid 3 = 3d

To calculate the relative density of a mixture, we use the equation:

\rho_{mix}=\frac{m_1+m_2+m_3}{V_1+V_2+V_3}ρ

mix

=

V

1

+V

2

+V

3

m

1

+m

2

+m

3

where,

m_1\text{ and }V_1m

1

and V

1

are the mass and volume of liquid 1

m_2\text{ and }V_2m

2

and V

2

are the mass and volume of liquid 2

m_3\text{ and }V_3m

3

and V

3

are the mass and volume of liquid 3