Question

When 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 the specific gravities of two liquids.

Answer 1

Hint: If the specific gravities of the two substances be ${S_1}$ and ${S_2}$, the density be $\rho $ , and $V$ be the volume of each of the substances, then, total mass of the mixture is $\left( {V{S_1}\rho + V{S_2}\rho } \right)$ . The specific gravity of the mixture is $\dfrac{{{S_1} + {S_2}}}{2} = 4$ and the specific gravity of the mixture is $\dfrac{{2{S_1}{S_2}}}{{{S_1} + {S_2}}} = 3$ . Simplify the two equations.
Formula used: If the specific gravities of the two substances be ${S_1}$ and ${S_2}$ , the density be $\rho $ , and $V$ be the volume of each of the substance, then, the total mass of the mixture is $\left( {V{S_1}\rho + V{S_2}\rho } \right)$ .
If, $m$ be the mass of each of the two substances, the total volume of the mixture is $\left( {\dfrac{m}{{{S_1}\rho }} + \dfrac{m}{{{S_2}\rho }}} \right)$ .

Complete step by step solution: Let the specific gravities of the two substances be ${S_1}$ and ${S_2}$ , and the density of water at $4^\circ C$ be $\rho $ .
So, densities of the substances are ${S_1}\rho $ and ${S_2}\rho $ .
Let $V$ be the volume of each of the substances in the mixture.
So, the total mass of the mixture is $\left( {V{S_1}\rho + V{S_2}\rho } \right)$ .
And, the total volume of the mixture is $\left( {V + V} \right) = 2V$ .
So, according to the problem, the specific gravity of the mixture is
$\dfrac{{{S_1} + {S_2}}}{2} = 4$
or, ${S_1} + {S_2} = 8$ $ \ldots \left( 1 \right)$
Again, let $m$ be the mass of each of the two substances in the mixture.
So, the total mass of the mixture is $\left( {m + m} \right) = 2m$ .
And, the total volume of the mixture is $\left( {\dfrac{m}{{{S_1}\rho }} + \dfrac{m}{{{S_2}\rho }}} \right)$ .
So, the density of the mixture is $\dfrac{{2m}}{{\dfrac{m}{{{S_1}\rho }} + \dfrac{m}{{{S_2}\rho }}}} = \dfrac{{2{S_1}{S_2}\rho }}{{{S_1} + {S_2}}}$
So, according to the problem, the specific gravity of the mixture is
$\dfrac{{2{S_1}{S_2}}}{{{S_1} + {S_2}}} = 3$
or, ${S_1}{S_2} = \dfrac{3}{2}\left( {{S_1} + {S_2}} \right) = \dfrac{3}{2} \times 8 = 12$
So, $\left( {{S_1} - {S_2}} \right) = \sqrt {{{\left( {{S_1} + {S_2}} \right)}^2} - 4{S_1}{S_2}} = \sqrt {{{\left( 8 \right)}^2} - \left( {4 \times 12} \right)} = 4$ $ \ldots \left( 2 \right)$
From equation $\left( 1 \right)$ , ${S_1} = 8 - {S_2}$
Putting this value in equation $\left( 2 \right)$ , we get, ${S_2} = 2$ .
And, putting this value in either of the equations, we get, ${S_1} = 6$ .

Note: The specific gravity of a solid or a liquid is defined as the ratio between the mass of a certain volume of the substance to the mass of an equal volume of water at $4^\circ C$ . Since the specific gravity of a substance is the ratio of two densities, it is also called relative density. Specific gravity is only a number and has no unit. For this reason, the value of the specific gravity of a substance is equal in all systems of units. It should be noted that it is a dimensionless physical quantity.