Question

When two equal volumes of two liquids are mixed, the specific gravity of the mixture is 4. When equal masses of the same two liquids are mixed the specific gravity of the mixture is 3. Find specific gravities of two liquids.

Answer 1

Hint:Determine the mass and volume of the two liquids for both the cases. From these values of the mass and volume of the mixture of liquids, derive the formula for the specific gravity of the mixture of the liquids and then solve the two equations obtained from these derivations to determine the values of specific gravities of the two liquids.

Formula used:
The density \[\rho \] of an object is given by
\[\rho = \dfrac{m}{V}\] …… (1)
Here, \[m\] is the mass of the object and \[V\] is the volume of the object.

Complete step by step answer:
We have given that the specific gravity of the mixture when equal volumes of the two liquids are mixed is 4.Let \[V\] be the volume of both liquids which are to be mixed. Let \[{m_1}\] and \[{m_2}\] be the masses of the two liquids and and \[{S_1}\], \[{S_2}\] be the specific gravities of the two liquids.The mass of this mixture is
\[{m_1} + {m_2} = {S_1}V + {S_2}V\]
The volume of the mixture is \[2V\].Hence, the specific gravity of the mixture according to equation (1) is
\[\dfrac{{{S_1}V + {S_2}V}}{{2V}} = 4\]
\[ \Rightarrow {S_1} + {S_2} = 8\] …… (2)

We have given that the specific gravity of the mixture when equal masses of the two liquids are mixed is 3.Let \[m\] be the mass of both liquids which are to be mixed. Let \[{m_1}\] and \[{V_2}\] be the volumes of the two liquids.The mass of this mixture is \[2m\].The volume of the mixture is
\[{V_1} + {V_2} = \dfrac{m}{{{S_1}}} + \dfrac{m}{{{S_2}}}\]
Hence, the specific gravity of the mixture according to equation (1) is
\[\dfrac{{2m}}{{\dfrac{m}{{{S_1}}} + \dfrac{m}{{{S_2}}}}} = 3\]
\[ \Rightarrow \dfrac{1}{{{S_1}}} + \dfrac{1}{{{S_2}}} = \dfrac{2}{3}\]
\[ \Rightarrow \dfrac{{{S_1} + {S_2}}}{{{S_1}{S_2}}} = \dfrac{2}{3}\] …… (3)

Substitute \[8 - {S_1}\] for \[{S_2}\] in the above equation.
\[ \Rightarrow \dfrac{{\left( {8 - {S_1}} \right) + {S_2}}}{{\left( {8 - {S_1}} \right){S_2}}} = \dfrac{2}{3}\]
\[ \Rightarrow \dfrac{8}{{8{S_1} - S_1^2}} = \dfrac{2}{3}\]
\[ \Rightarrow 16{S_1} - 2S_1^2 = 24\]
\[ \Rightarrow S_1^2 - 8{S_1} - 12 = 0\]
\[ \Rightarrow {S_1} = 6,2\]
Substitute \[6\] for \[{S_1}\] in equation (2).
\[ \Rightarrow 6 + {S_2} = 8\]
\[ \therefore {S_2} = 2\]

Hence, the specific gravities of the two liquids are 6 and 2.

Note:The students should keep in mind that we have used the formula of the density of an object for reference to derive the relation for specific gravity of the mixture of two liquids. But the specific gravity of a substance is not the normal density. It is the ratio of actual density to density of the reference material.