## Introduction

In this post, I will show you a very simple way on how to calculate the relative density of a liquid. I will explain the steps you need to take in order to solve problems on the relative density of liquids.

You may also like to read:

## Explanation on How to Calculate the Relative Density of a Liquid

Here I will explain two steps you can apply, so that you will understand how to calculate the relative density of a liquid

### Step 1

I will use an empty dry bottle to explain in simple terms, how to calculate the relative density of a liquids. Now, lets get started:

- Get a dry empty bottle, record it is mass and name it M
_{1}- Add liquid (for example oil) to the bottle and name it M
_{2}. Now, mass of the liquid is added to the mass of the mass of the empty bottle and it is M_{2}.- The next step is to remove the liquid from the bottle.
- Wash and dry the bottle again
- Add water to the bottle, and record their total mass together. Tag this mass M
_{3}

To Find only the mass of the liquid without the bottle, subtract **Mass of the empty bottle (M _{1})** from

**Mass of the bottle and the liquid (M**which is

_{2})**Mass of liquid only = M _{2} – M_{1}**

Additionally, you need to apply the following methods to find the **Mass of an equal volume of waterequal volume of water:**

subtract **Mass of the empty bottle (M _{1})** from

**Mass of the bottle and that of the water together (M**

_{3}

**)**which is:

**Mass of equal volume of water** **= M**_{3}** – M _{1}**

Therefore, we can now say that the formula for how to calculate relative density of a liquid is

**Relative density of liquid = Mass of the liquid / Mass of equal volume of water**

which is

**Relative density of liquid =** **(M _{2} – M_{1})**

**/ (M**

_{3}

**– M**

_{1})### Step 2

In this step, I will apply Archimedes principle to explain how to calculate the relative density of liquid:

- Attach a metal object to spring balance and suspend it in air. Record the
mass of the object in air as M._{1}- Immerse the metal object (that is already attached to the spring balance) in liquid, and record the
apparent mass of liquid as M_{2}- Clean and dry the metal object
- Immerse the object in the water and record the
apparent mass of water as M_{3}- We will now find the weight of the liquid that is being displaced by the metal object, which is the upthrust in liquid.
- Now, the
upthrust in liquidsis calculated by subtractingmass of the object in air as Mfrom_{1}apparent mass of liquid as M. This implies that_{2}Upthrust in liquids=(M_{2}–M)_{1}- Upthrust in water is the same thing as the weight of the water displaced by the object
- To find upthrust in water, we subtract
apparent mass of water (Mfrom_{3})mass of the object in air (M. Therefore,_{1})Upthrust in water =M–_{1}M_{3}- Hence, to calculate Relative density, we say that

Relative density =[Upthrust in liquids=M_{2}–M_{1}]/[Upthrust in water =M–_{1}M_{3}]Relative density =(M_{2}–M)_{1}/(M–_{1}M_{3})

## Examples of How to Calculate the Relative Density of Liquids

Remember to follow the following steps in order to solve your examples at ease:

- Underline all the figures you see in the question
- Read the question, and extract your data
- Underline the statement where you see question and write the formula you need to apply
- Check all the elements in the formula and assign their figures to them
- Solve the problem by making the element in question subject of the formula
- Don’t forget to always include the s.i unit in your final answer

### Example 1

A body weighs 0.30N in air, 0.25N when fully immersed in water, and 0.27N when fully immersed in a liquid. Calculate (a) it is loss of weight in water, (b) relative density, and (c) The relative density of the liquid

**Solution:**

(a) To calculate the loss of weight in water, we subtract the **weight of the body when fully immersed in water (0.25N)** from the **weight of the body in air (0.30N)**.

This implies that

**Loss of weight in air = 0.30 – 0.25 = 0.05N**

Therefore,

**Loss of weight in air = Upthrust in water = Weight of equal volume of water displaced by the body = 0.05N**

(b) **Relative density of the body = Weight of the body in air (0.30) / Weight of equal volume in water (0.05)**

Now, we can see that

**Relative density of the body =** **0.30 / 0.05** **=** **6**

(c) Relative density of liquid = Upthrust in liquid (0.30 – 0.27) / Upthrust in water (0.30 – 0.25)

This is because

Upthrust in liquid = mass of the body in air – mass of the body in liquid

and Upthrust in water = mass of the body in air – mass of the body in water

Thus,

Relative Density of the liquid = (0.30 – 0.27) / (0.30 – 0.25)

Therefore, the Relative density of liquid = 0.03 / 0.05 = 0.6

Now we can see that Relative density of liquid is 0.6

### Example 2

A relative density bottle weighs 20 grams when empty, 80 grams when filled with water, and 100 grams when filled with a liquid. Find the relative density of the liquid.

**Solution:**

**Data**

Weight of the bottle in air, W_{1} = 20 grams

Also, the weight of the bottle with water, W_{2} = 80 grams

Additionally, weight of the bottle with liquid, W_{3} = 100 grams

Applying the formula

Relative density = (W_{3} – W_{1}) / (W_{2} – W_{1})

Thus, we have

Relative density = (100 – 20) / (80 – 20) = 80 / 60 = 1.33

### Example 3

A density bottle has a mass of 0.05 kilogram when empty, a mass of 0.30 when some quantity of steel ball bearings is added to it and a mass of 0.35 kilogram when the remainder of the bottle is filled with water. if the density of the bottle weighs 0.2 kilogram when filled with water. Calculate the relative density of the steel ball bearing.

Solution:

Data

Mass of an empty bottle, M_{1} = 0.05 kg

Also, Mass of bottle with steel ball bearings, M_{2} = 0.30 kg

Mass of the bottle, steel ball bearings, and water, M_{3} = 0.35

we can also see that Mass of the bottle with water, M_{4} = 0.2 kg

The Relative density of the steel (R.D) = ?

Now, let us apply the formula relative density (r.d) to solve the above problem

Mass of the steel ball bearing M_{5} = M_{2} – M_{1} = 0.30 – 0.05 = 0.25 kg

We can also see that

Mass of water and the bottle M_{6 }= M_{4} – M_{1} = 0.2 – 0.05 = 0.15 kg

Additionally, mass of water, the bottle, and the steel ball bearing M_{7 }= M_{3} – M_{2 = }0.35 – 0.30 = 0.05 kg

We finally calculate the mass of water having the same volume as the ball bearing M_{8} = M_{6} – M_{7} = 0.15 – 0.05 = 0.10

Therefore to find the relative density of steel ball bearing, we say that

Relative density (R.d) = M_{5} / M_{8} = 0.25 / 0.10 = 2.5

### Example 4

The mass of a bottle is 18 gram when it is empty, 44.0 gram when it is filled with water and 39.84 gram when it is full of a second liquid. Calculate the density of the liquid.

Solution:

Data:

Mass of an Empty bottle M_{1 }= 18.0 g

Also,

Mass of the bottle and that of water M_{2} = 44.0 g

Therefore, we have mass of water, M_{3} = M_{2} – M_{1} = 44.0 – 18.0 = 26 g

Also, mass of the bottle and the liquid M_{4} = 39.84 g

Now, mass of liquid, M_{5} = M_{4} – M_{1} = 39.84-18.00 = 21.84 g

The relative density of the liquid = mass of liquid / mass of equal volume of water = M_{5} / M_{3}

R.d = 21.84/26 = 0.84

The density of the liquid = 0.84 g/cm^{3}

**Note:**

Relative density is simply comparing the density of substances (solid, liquid or gaseous) to that of water (1 g/cm^{3}).It does not have a unit and may be measured with a hydrometer. We use the symbol R.D to indicate relative density, and its a scalar quantity

You may also like to read:

A Sample of Physics Lesson Plan