Two liquids of densities d1 and d2 are mixed in equal masses. Find the resultant density of the mixture.

Question

Prepp Team · Accepted Answer

Understanding Density of a Mixture

This question asks us to find the resultant density when two liquids with different densities are mixed in equal masses. Density is a fundamental property of a substance, defined as its mass per unit volume.

Let's denote the densities of the two liquids as $d_1$ and $d_2$. We are told that the two liquids are mixed in equal masses. Let this equal mass be $m$.

Calculating Volumes

The relationship between mass ($m$), density ($\rho$), and volume ($V$) is given by:

$\rho = \frac{m}{V}$

From this, we can express volume as:

$V = \frac{m}{\rho}$

For the first liquid, with mass $m_1 = m$ and density $d_1$, the volume is:

$V_1 = \frac{m_1}{d_1} = \frac{m}{d_1}$

For the second liquid, with mass $m_2 = m$ and density $d_2$, the volume is:

$V_2 = \frac{m_2}{d_2} = \frac{m}{d_2}$

Determining Total Mass and Total Volume

When the two liquids are mixed, the total mass of the mixture ($M$) is the sum of the individual masses:

$M = m_1 + m_2 = m + m = 2m$

Assuming the volumes are additive (no change in volume upon mixing), the total volume of the mixture ($V$) is the sum of the individual volumes:

$V = V_1 + V_2 = \frac{m}{d_1} + \frac{m}{d_2}$

We can factor out $m$ from the expression for total volume:

$V = m \left( \frac{1}{d_1} + \frac{1}{d_2} \right)$

To combine the fractions inside the parenthesis, we find a common denominator ($d_1 d_2$):

$V = m \left( \frac{d_2}{d_1 d_2} + \frac{d_1}{d_1 d_2} \right) = m \left( \frac{d_1 + d_2}{d_1 d_2} \right)$

Calculating the Resultant Density

The resultant density of the mixture ($d$) is the total mass divided by the total volume:

$d = \frac{M}{V}$

Substitute the expressions for $M$ and $V$:

$d = \frac{2m}{m \left( \frac{d_1 + d_2}{d_1 d_2} \right)}$

The term $m$ cancels out from the numerator and denominator:

$d = \frac{2}{\frac{d_1 + d_2}{d_1 d_2}}$

To simplify, we invert the denominator and multiply:

$d = 2 \times \frac{d_1 d_2}{d_1 + d_2}$

$d = \frac{2 d_1 d_2}{d_1 + d_2}$

This formula gives the resultant density when two liquids are mixed in equal masses.

Comparing with Options

Let's compare our derived formula with the given options:

Option 1: $\frac{{{d_1}{d_2}}}{{{d_1} + {d_2}}}$
Option 2: $\sqrt {{d_1}{d_2}} $
Option 3: $\frac{{{d_1} + {d_2}}}{2}$
Option 4: $\frac{{2{d_1}{d_2}}}{{{d_1} + {d_2}}}$

Our derived resultant density is $\frac{{2{d_1}{d_2}}}{{{d_1} + {d_2}}}$, which matches Option 4.

Quantity	Liquid 1	Liquid 2	Mixture
Density	$d_1$	$d_2$	$d$ (Resultant)
Mass	$m$	$m$	$M = 2m$
Volume	$V_1 = \frac{m}{d_1}$	$V_2 = \frac{m}{d_2}$	$V = V_1 + V_2 = m\left(\frac{1}{d_1} + \frac{1}{d_2}\right)$
Formula	$d_1 = \frac{m}{V_1}$	$d_2 = \frac{m}{V_2}$	$d = \frac{M}{V} = \frac{2m}{m\left(\frac{1}{d_1} + \frac{1}{d_2}\right)} = \frac{2}{\frac{d_1+d_2}{d_1d_2}} = \frac{2d_1d_2}{d_1+d_2}$

Revision Table: Liquid Mixture Density

Understanding how density changes upon mixing is crucial. The formula for the mixture density depends on whether the liquids are mixed by mass or by volume.

Mixing by Equal Masses: If mass $m$ of liquid 1 (density $d_1$) and mass $m$ of liquid 2 (density $d_2$) are mixed, the resultant density $d$ is given by $d = \frac{2d_1d_2}{d_1+d_2}$. This is the case solved in this problem.
Mixing by Equal Volumes: If volume $V$ of liquid 1 (density $d_1$) and volume $V$ of liquid 2 (density $d_2$) are mixed, the total volume is $2V$. The mass of liquid 1 is $m_1 = d_1V$, and the mass of liquid 2 is $m_2 = d_2V$. The total mass is $M = d_1V + d_2V = (d_1+d_2)V$. The resultant density $d$ is $d = \frac{M}{2V} = \frac{(d_1+d_2)V}{2V} = \frac{d_1+d_2}{2}$. This is simply the arithmetic mean of the densities.

Additional Information: Average Density Concepts

The resultant density when mixing substances is a type of average density. When mixing by mass, the resultant density formula $\frac{2d_1d_2}{d_1+d_2}$ is related to the harmonic mean. Specifically, the reciprocal of the resultant density is the arithmetic mean of the reciprocals of the individual densities:

$\frac{1}{d} = \frac{V}{M} = \frac{V_1+V_2}{m_1+m_2}$

If masses are equal ($m_1=m_2=m$) and volumes are $V_1=m/d_1$, $V_2=m/d_2$:

$\frac{1}{d} = \frac{m/d_1 + m/d_2}{m + m} = \frac{m(1/d_1 + 1/d_2)}{2m} = \frac{1}{2} \left( \frac{1}{d_1} + \frac{1}{d_2} \right)$

$\frac{1}{d} = \frac{1}{2} \left( \frac{d_2 + d_1}{d_1 d_2} \right)$

$\frac{1}{d} = \frac{d_1 + d_2}{2 d_1 d_2}$

Taking the reciprocal of both sides:

$d = \frac{2 d_1 d_2}{d_1 + d_2}$

This confirms the result using an alternative perspective based on the reciprocal of density (specific volume if mass is unity).

Prepp Team · Answer

$\frac{{{d_1}{d_2}}}{{{d_1} + {d_2}}}$

Prepp Team · Answer

$\sqrt {{d_1}{d_2}} $

Prepp Team · Answer

$\frac{{{d_1} + {d_2}}}{2}$

Quantity	Liquid 1	Liquid 2	Mixture
Density	\(d_1\)	\(d_2\)	\(d\) (Resultant)
Mass	\(m\)	\(m\)	\(M = 2m\)
Volume	\(V_1 = \frac{m}{d_1}\)	\(V_2 = \frac{m}{d_2}\)	\(V = V_1 + V_2 = m\left(\frac{1}{d_1} + \frac{1}{d_2}\right)\)
Formula	\(d_1 = \frac{m}{V_1}\)	\(d_2 = \frac{m}{V_2}\)	\(d = \frac{M}{V} = \frac{2m}{m\left(\frac{1}{d_1} + \frac{1}{d_2}\right)} = \frac{2}{\frac{d_1+d_2}{d_1d_2}} = \frac{2d_1d_2}{d_1+d_2}\)