Question

What is the volume of a solution that has a specific gravity of $1.2$ and mass of $185\,g$?

Answer 1

Hint: In order to solve this question, we will be using the formulae for the specific gravity of a solution, and we will also look at the definition of specific gravity of a solution.

Formula used:
The formula for specific gravity of a solution
$sp.gravity=\dfrac{\rho (solution)}{\rho (water)}$
Where $\rho $ is the density.

Complete step by step answer:
First of all, we will write down the definition of specific gravity and understand where the formula mentioned above came from. For this purpose, we will be using an example. When we go out in the summer to enjoy the pools and rivers, we will always double-check that all of our equipment and safety precautions are in place and operational. When we go to the river and board our vessels, we make sure there are no leaks and that we are wearing our life jackets.

We make sure that young children who are either learning to swim have their floaters on or swim with them in a swim ring to ensure that they do not sink. Why do both vessels and swimming rings float? One of the variables that determines whether an object can sink or float in water is what is known as the specific gravity.The specific gravity of an object is the ratio of its mass to that of a reference material. The specific gravity will tell us whether an object would sink or float in our reference substance depending on its meaning.

Water is commonly used as a reference substance since it has a density of one gram per milliliter or one gram per cubic centimeter. If it is provided that the reference material is water, the formula for specific gravity is the object's mass divided by the density of the water.The Greek letter Rho is used to denote density in this case.
$sp.gravity=\dfrac{\rho (solution)}{\rho (water)}$

The specific gravity of the solution$sp.gravity=1.2$
The mass of the solution$m=185g$
Using this formula, we will find the density of the solution first.
$\text{sp.gravity}=\dfrac{\rho (solution)}{\rho (water)} \\
\Rightarrow \rho (solution)=sp.gravity\times \rho (water) \\
\Rightarrow \rho (solution)=1.2\times 1\dfrac{g}{ml} \\
\Rightarrow \rho (solution)=1.2\dfrac{g}{ml}$

As we know the density of water is $1\dfrac{g}{ml}$ and the specific gravity is unitless quantity. Now, we can find the volume of the solution from the density as,
$\text{density}=\dfrac{mass}{volume} \\
\Rightarrow \text{volume}=\dfrac{mass}{density} \\
\Rightarrow \text{volume}=\dfrac{mass}{\rho (solution)} \\
\Rightarrow \text{volume}=\dfrac{185g}{1.2\dfrac{g}{ml}} \\
\therefore \text{volume}=154\,ml$

Hence, the volume of a solution is 154 ml.

Note: The density of water that is used above is at ${{4}^{0}}C$ which is considered as standard. If the temperature of water is specified, then use the density at that particular temperature.