Question

Three liquids of densities $d,{\text{ }}2d{\text{ }}and{\text{ }}3d$ are mixed in equal proportions of weights. The relative density of the mixture is
$A)\dfrac{{11d}}{{7}}$
$B)\dfrac{{18d}}{{11}}$
$C)\dfrac{{13d}}{9}$
$D)\dfrac{{23d}}{{18}}$

Answer 1

According to the question, three liquids of densities $d,{\text{ }}2d{\text{ }}and{\text{ }}3d$ are mixed in equal proportions of weights. So, here we need to calculate the relative density of the mixture. To calculate this, we have the formula ${D_{mix}} = \dfrac{{{M_1} + {M_2} + {M_3}}}{{{V_1} + {V_2} + {V_3}}}$

Complete Step By Step Answer:
A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a constant volume independent of pressure. As such, it is one of the four fundamental states of matter, and is the only state with a definite volume but no fixed shape. The density of a substance is the relationship between the mass of the substance and how much space it takes up (volume). The mass of atoms, their size, and how they are arranged determine the density of a substance. Density equals the mass of the substance divided by its volume, $D = \dfrac{M}{V}$ .
The difference between the specific gravity and density is that at room temperature and pressure is $1$ gram per $1$ cubic cm is the density of water this density is treated as a standard and the density of any other material (usual liquids) is calculated relative to this is called relative density or specific gravity. Another way to define Relative density is that it is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material (i.e., water). It is usually measured at room temperature ( $20$ Celsius degrees) and standard atmosphere $\left( {101.325kPa} \right)$
We are given,
The density of liquid $1{\text{ }} = {\text{ }}d$
The density of liquid $2{\text{ }} = {\text{ }}2d$
The density of liquid $3{\text{ }} = {\text{ }}3d$
Since the mass of all the liquids are same,
$Mass{\text{ }}of{\text{ }}liquid{\text{ }}1{\text{ }} = {\text{ }}Mass{\text{ }}of{\text{ }}liquid{\text{ }}2{\text{ }} = {\text{ }}Mass{\text{ }}of{\text{ }}liquid{\text{ }}3{\text{ }} = {\text{ }}m$
Now,
The volume of liquid $1$ = $\dfrac{{{m_1}}}{{{D_1}}}$
The volume of liquid $2$ = $\dfrac{{{m_2}}}{{{D_2}}}$
The volume liquid $3$ = $\dfrac{{{m_3}}}{{{D_3}}}$
So, the relative density of the mixture is = $\dfrac{{m + m + m}}{{\dfrac{m}{D} + \dfrac{m}{{2D}} + \dfrac{m}{{3D}}}}$ ,
Now by cancelling the numerator by the denominator, we get,
= $\dfrac{{3 \times 6D}}{{11}}$ = $\dfrac{{18D}}{{11}}$
Thus, the final answer is $\dfrac{{18D}}{{11}}$ .

Additional Information:
Relative density is often used to calculate the volume or weight of samples needed for preparing a solution with a specified concentration. It also helps us understand the environmental distribution of insoluble substances (i.e., oil spill) in aquatic ecosystems (on water surface or bottom sediment) if the substance is released to water.

Note:
The difference between the specific gravity and density is that at room temperature and pressure is 1 gram per 1 cubic cm is the density of water this density is treated as a standard and the density of any other material (usual liquids) is calculated relative to this is called relative density or specific gravity.