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LEARNING OBJECTIVES By the end of this chapter, you should be able to: 1. Define: – Density and state its S.I unit and the other unit. 2. Determine experimentally the densities of: – Regular and Irregular Solids, – Liquids and – Gases/Air. 3. Solve numerical problems on density and density of mixtures. 4. Define: – Relative Density (R.D) – Determine experimentally the R.D of; – Solids and – Liquids. 5. Solve numerical problems on relative density of solids and liquids. |
3.0 Density
Definition: Density is defined as mass per unit volume of a substance.
Mathematically, it is expressed as: Density =
ρ =
(a) S.I Unit
The SI unit of density is kg/m^{3} (kgm^{-3}). It is a derived unit. I.e. a unit derived from the units of the quantities in the formula of density.
(b) Derivation of the unit
From Density = we have;
=
Conveniently expressed as = or
The smaller unit of density is g/cm^{3} (g cm^{-3}).
Note: The density of a substance is subject to variation depending on the prevailing physical factors.
3.1 (a) Factors that affect density
There are two physical factors that affect the density of a substance. These are:
- Temperature and Pressure.
(i) Effect of Temperature on density of substances
Substances expand and contract when their temperatures changes. The expansion and the contraction cause increase and decrease in the volume. Since, density is the ratio of mass to volume of a substance and then there will be change in the value of the density.
Effect of High temperature
At high temperature, a substance absorbs heat energy and expands and the volume is increased. Since the mass remains constant, the value of the density (i.e. the ratio of mass to volume) becomes low i.e. the density becomes low.
Effect of low temperature
At low temperature, a substance loses heat energy and contracts. The volume decreases. Dividing the constant mass by the reduced volume gives a higher density than the density under normal conditions.
(ii) Effect of Pressure on Density of substances
Pressure as a determining factor of density mostly affects the density of gases.
High pressure
High pressure squeezes the gas molecules into a smaller volume. If a given mass of a gas is contained within a smaller volume, the ratio of mass to volume results to a higher density.
Low pressure
At low pressure, the gas molecules occupy a larger volume and therefore, the density becomes low.
Note: As a result of the above factors, when stating densities of gases, the temperature and pressure values must be stated as well.
The table 3.1 below shows the densities of some common substances/ materials.
Substance | Density in kg m^{-3} |
(a) Solids: – Ice | 920 (9 x 10^{2}) |
– Aluminium | 2,700 (2.7 x 10^{3}) |
– Copper | 8,900 (8.9 x 10^{3}) |
– Lead | 11,300 (1.13 x 10^{4}) |
(b) Liquids: – Water | 1,000 (1 x 10^{3}) |
– Paraffin, gasoline | 800 (8 x 10^{2}) |
– Mercury | 13,600 (1.36 x 10^{4}) |
Gases (at *stp) – Air | 1.30 |
– Hydrogen | 0.09 |
– Oxygen | 1.43 |
– Carbon dioxide | 1.98 |
Table 3.1
Note: (i) The above density values are not be memorised except for water.
(ii) *stp – stands for Standard Temperature and Pressure.
(b) Uses of Density
Density is used to:
(i) Identify materials.
(ii) Determine the purity of a material.
(iii) Choose light gases for filling meteorological balloons.
(c) Importance of density
The densities of materials are important to architectures and engineers in the design of structures. For example, aircrafts and overhead cables for the transmission of electricity are made of aluminium alloy. This is because aluminium has low density (i.e. is light) and is quite strong.
Worked Examples
Calculating density
Steps in problem solving
Before solving any problem, ask your self the following questions.
- What is asked in the question?
- What information is given to help solve the problem?
- What are the equation(s) to solve the problem?
- Are units of the quantities given matching?
These questions can only be answered when you collect the data.
1. A glass stopper has a volume of 16 cm^{3} and a mass of 40 g. Calculate the density of
the glass stopper in: (i) g/cm^{3 }(ii) kg/m^{3}
Solution
Data: m = 40 g, v = 16 cm^{3}, ρ = ?
(i) Density = = = 2.5 g/cm^{3}
Note: For (ii), the units of mass and volume in the data are small units. But you are required to get the answer in kg/m^{3}. This means that the mass must be in kg and the volume in m^{3}. So first convert the mass from gram to kg and volume from cm^{-3} to m^{3}.
Converting the units: (ii) Mass: 40g = = 0.04 kg = 4.0 x 10^{-2} kg
Vol; 16 cm^{3 }= = 0.000016 = 1.6 x 10^{-6} m^{3}
Now calculate the density: ρ = =
= 2.5 x 10^{3} kg/m^{3}
2. The mass of 24.4 cm^{3} of mercury is 332 g. Find the density of mercury.
Solution: Data: m = 166 g, v = 12.2 cm^{3}, r = ?
r = = = 13.6 g/cm^{3}
3. 25 cm^{3} of aluminium has a mass of 67.5 g. Calculate its density in
(a) g/cm^{3} (b) kg/m^{3}
Solution: Data: (a) v = 25 cm^{3 }, m = 67.5 g, r = ?
r = =
= 2.7 g/cm^{3 }
(b) First convert the mass and the volume into their respective SI units.
Þ Converting the
mass: 1 kg = 1 000 g
m = 67.5g
=
\ m = 0.0675 kg
Þ Converting the
volume:1 m^{3 }= 1 000 000 cm^{3}
v = 25 cm^{3}
=
\ v = 0.000 025 m^{3}
Now applying the formula
r = we have;
=
\
r = 2700 kg/m^{3} or 2.7 x 10^{3} kgm^{-3}
Calculating mass
4. The density of copper is 8.9 g/cm^{3}. What is the mass of 100 cm^{3} of copper?
Solution: Data: r = 8.9 g/cm^{3}, v = 100 cm^{3}, m = ?
r =
8.9 =
m = 8.9 x 100
\ m = 890 g
Calculating volume
5. Calculate the volume of a block of expanded polystyrene of mass 400 g if its density is 16 kg/m^{3}.
Solution Data: m = 400 g, ρ = 16 kg/m^{3}, v = ?
Note: The mass must be converted to kilograms to match the density unit.
m = = 0.4 kg
Rearranging the formula for calculating density we have:
v = = = 0.025 m^{3}
3.2 Measurement of Density
(a) To find the density of a Regular Solid
- Measure the dimensions of the solid object i.e. length, width, height or diameter, using an appropriate instrument.
- Calculate the volume of the object from the appropriate formula.
Say volume of object = y m^{3}
- Find the mass of the object using a triple balance. Say mass = x kg.
- Calculate the density from the formula
Density =
\ Density =
Examples
1. A cuboid of wood of mass 20 g measures 5 cm by 4 cm by 2 cm. Find its density.
Solution: Data: m = 20 g, l = 5 cm, w = 4 cm, h = 2 cm, r = ?
Notice from the data that mass is given and volume is not given. Therefore, we first get the volume by substituting the dimensions in the formula of volume of a cuboid.
v = lwh = 5 x 4 x 2 = 40 cm^{3}
Now calculate the density, r = = = 0.5 g/cm^{3}
2. A spherical metal made of aluminium weighs 90.477 g in air. If the diameter of the sphere is 4.0 cm^{3}, find the density of the sphere. (Take p = 3.14)
Solution: Data: m = 90.477 g, d = 4.0 cm (r = 2 cm), v = ?, r = ?
v = pr^{3 }= = = 33.49 cm^{3}^{ }
Applying the formula r =
= = 2.7 g/cm^{3}
(b) To find the density of an Irregular
Solid
- Pour water in a measuring cylinder and record the first reading of the water level, say x cm^{3}
- Tie the irregular solid with a piece of thin silk thread and carefully immerse it into the water in the measuring cylinder.
- Read and record the second reading of the water level, say y cm^{3}.
- Find the volume of the irregular solid from the formula:
Volume of object = (Second reading – First reading)
= (y – x) cm^{3}
– Determine the mass of the solid on a triple balance, say mass = z g.
Calculation;
Results: Mass of object in air = z g
Volume of object = (y – x) cm^{3}
Using the formula
Density = = =
Example
When a piece of irregular stone of mass 164.5 g was immersed in 300 cm^{3} of water in a measuring cylinder, the level of water rose to 370 cm^{3}. Calculate the density of the stone.
Solution: Data: Density of stone, r
= ?
Mass of stone in air = 164.5 g
Initial reading of water = 300 cm^{3}
Final reading = 370 cm^{3}
Volume of stone = Final reading – Initial reading
= 370 – 300
= 70 cm^{3}
Density =
=
\
r = 2.35 g/cm^{3}
- To find the density of liquid e.g. Paraffin
Procedure
- Weigh an empty beaker on a triple balance, say x grams.
- Pour a known volume, v, of the liquid in the beaker.
- Weigh the beaker and the liquid (paraffin). Let the total mass be y grams.
- Calculate the density as below:
Results:
Mass of empty beaker = x g
Mass of beaker + Paraffin = y g
Mass of paraffin only = (y – x) g
Volume of paraffin = v cm^{3}
Density =
=
Example
An empty beaker weighs 120 g in air. And weighs 180 g when filled with 75 cm^{3 }of methylated spirit. Find the density of the methylated spirit.
Solution: Mass of empty beaker = 120 g
Mass of beaker + Paraffin = 180 g
Mass of paraffin only = (180 – 120) g
Volume of paraffin = 75 cm^{3}
Density =
=
\ Density of methylated spirit = 0.8 gcm^{-3}
3.2 Density of Mixtures
A mixture is a substance that consists of two or more substances physical combined together.
Mixtures are obtained by mixing two or more substances physically. In dealing with the calculations of density of mixtures, the following assumptions are made.
- The constituents of the mixture do not react with one another.
- The total mass of the mixture is the sum of the masses of the constituents.
The density of mixtures is calculated from the formula.
Density of mixture =
N.B (a) The density of the mixture lies between the densities of its constituents.
(b) Calculations on density of mixtures are of two types.
- Where the masses and volumes of the constituents are given directly.
(ii) Where either the masses and densities are given but volumes not given.
Or volumes and densities are given but masses not given.
In the case of the (b) (i), the formula for calculating density of mixture is applied directly after getting the total mass and total volume of the mixture.
While for the case of (b) (ii), we first use the formula for calculating density to get the quantities which are not given. There after we apply the formula of density of mixtures.
(a) Calculating density of a mixture when the masses and the volumes of the
constituents are given
Example 1
100 cm^{3} of fresh water which weighs 100g is mixed with 100cm^{3} of sea water which weighs 103g .Calculate the density of the mixture?
Hint: For this type of question, we get the total mass and total volume of the mixture and then substitute the values in the formula of density of mixtures. See examples below.
Solution: Mass of fresh water = 100 g
Mass of sea water = 103 g
Mass of the mixture = 100 g + 103g
= 203 g
Volume of fresh water = 100 cm^{3}
Volume of sea water = 100 cm^{3}
Volume of the mixture = 100 cm^{3 }+ 100 cm^{3}
= 200 cm^{3}
Density of mixture =
=
\ Density of mixture = 1.015 g/cm^{3}
(b) Calculating density of a mixture when either masses or volumes of the constituents are not given
Example 2
0.0018 m^{3 }of fresh water of density 1000 kg/m^{3} is mixed with 0.0022 m^{3 }of sea water of density 1,025 kg/m^{3 }. Calculate the density of the mixture.
Solution: Mass of fresh water = ?
Volume of fresh water = 0.0018 m^{3}
Density of fresh water = 1000 kg/m^{3}
Mass of sea water = ?
Volume of sea water = 0.0022 m^{3}
Density of fresh water = 1,025 kg/m^{3}
Hint: Note that the masses are not given. So use the formula of density of a substance to get the masses of the constituents first and then follow the steps in example 1 above.
Þ Calculating mass of fresh water: ρ =
m = ρv
= 1000 x 0.0018
\ Mass of fresh water = 1.8 kg
Þ Calculating mass of sea water: ρ =
m = ρv
= 1000 x 0.0018
= 1.025 x 0.0022
\ Mass of fresh water = 2.255 kg
Since the masses and the volumes of the constituents are now known, the density of the mixture can calculated from the data below.
Density of the mixture
Mass of fresh water = 1.8 kg
Mass of sea water = 2.255 kg
Mass of the mixture = 1.8 + 2.255
= 4.055 kg
Volume of fresh water = 0.0018 m^{3}
Volume of sea water = 0.0022 m^{3}
Volume of the mixture = 0.0018 m^{3 }+ 0.0022 m^{3}
= 0.004 cm^{3}
Density of mixture =
=
\ Density of mixture = 1.013 kg/m^{3}
Note: The detailed steps are to make you to understand how to solve problems in this topic. But in examination you are required to present your work brief and clear.
3.3 RELATIVE DENSITY (R.D)
Definition
Relative density of a substance is defined as the ratio of the density of a substance to the density of water.
Mathematically, R.D is expressed as:
Relative density (R.D) =
If the masses of equal volume of a substance and water are found, then this relation takes the form
Relative density (R.D) =
In a normal weighing, the mass of a substance is proportional to the weight, so it is also true to say;
Relative density (R.D) =
NB: R.D has no unit since it is a ratio of the same quantity i.e. densities, masses or force (weigh)t as a result the units cancel.
3.31 Measurement of Relative Density (R.D)
(a) To measure the density of Liquid
The R.D of a liquid is measured by using a density bottle. The density bottle has glass stopper with a fine hole through it, so that, when it is filled fully with the liquid and the stopper inserted, the excess liquid rises through the fine hole and runs down the outside.
So long as the bottle is used to the same liquid level at the top of the hole, it will always contain the same volume of whatever liquid is filled in it provided the temperature remains the same.
Experiment to determine the R.D of liquid e.g. Paraffin
Procedure
- Weigh the density bottle when empty.
- Fill the bottle full with the paraffin.
- Wipe the paraffin that runs out through the hole and weigh the bottle and the paraffin.
- Empty the bottle and clean it thoroughly.
- Refill the bottle with water to the same level and weigh it after wiping the water that flows out.
Calculate the R.D as shown below.
Mass of empty bottle = x g
Mass of bottle full of liquid = y g
Mass of bottle full of water = z g
Mass of liquid = (y – x) g
Mass of water = (z – x) g
Applying the formula
Relative density of liquid =
=
\R.D =
(b) Precautions
To obtain accurate result, the following precautions should be taken.
(i) The outside of the bottle must be wiped dry before weighing.
(ii) The bottle must not be held by the neck with warm hand otherwise some liquid may be lost due to expansion.
Worked Examples
Steps in problem solving
Before solving any problem, ask your self the following questions.
- What is asked in the question?
- What information is given to help solve the problem?
- What are the equations to solve the problem?
These questions are answered when you collect the data.
Example 1
A density bottle was used to measure the relative density of a liquid and the following results were obtained.
Solution: Mass of empty density bottle = 30 g
Mass of bottle full of liquid = 110 g
Mass of bottle full of water = 130 g
Mass of liquid = 110 – 30
= 80 g
Mass of water = 130 – 30
= 100 g
Relative density of liquid =
=
\ R.D of liquid = 0.8
Example 2
The mass of an empty density bottle is 46.00 g. When fully filled with water it weighs 96 g. And when full of a liquid of unknown R.D. it weighs 86 g.
Calculate: (i) the R.D of the liquid.
(ii) the density of the liquid.
Solution: (i) Mass of empty bottle = 46 g
Mass of bottle full of liquid = 96 g
Mass of bottle full of water = 86 g
Mass of liquid = 86 – 46
= 40 g
Mass of water = 96 – 46
= 50 g
Relative Density of liquid =
=
\ R.D of liquid = 0.8
(ii) R.D of liquid = 0.8, density of water = 1000 kgm^{-3}
Relative density (R.D) =
0.8 =
Density of liquid = 0.8 x 1000
\ Density of liquid = 800 kgm^{-3}
(c) To measure the Relative Density of Solid
The R.D of solid/liquid substances can best be measured by applying Archimedes’s Principle to be discussed in detail later.
Archimedes’s Principle stats that:
When a body is wholly or partially immersed in a fluid, it experiences an upthrust equal to the weight of fluid displaced.
Procedure
- Suspend the solid whose relative is to be determined from a spring balance by means of light string in air and record its weight.
- Immerse the solid wholly in water and record its apparent weight.
Results:
Let: Weight of solid object in air = W_{a} N
Weight of solid object in water (Apparent weight) = W_{w} N
Calculation:
Upthrust (Loss in weight of object) = Weight of water displaced
= Weight in air – Apparent weight
\ Upthrust = (W_{a} – W_{w})_{ }N
But Volume of water displaced = Volume of the solid immersed
From Relative Density =
=
I.e. Relative Density (R.D) =
Note: For details on R.D of solids, see chapter 10.
Examples
1. A piece of aluminium weighs 80 N in air and 50.37 N when completely immersed in water. Calculate the relative density of glass.
Solution: (Remember the steps in problem solving!!)
Weight of solid object in air = 80 N
Weight of solid object in water (Apparent weight) = 50.37 N
Weight of water displaced = Upthrust in water = 80 – 50.37 N
= 29.63 N
Relative Density =
=
= 2.699966
\ Relative Density = 2.7
SELF-CHECK 3.0
1. To calculate the density of an object, which one of the following must be known?
I. Height II. Volume III Area IV Mass V Weight
A. I and II B. II and V. C. III and IV D. II and IV A
2. A block of wood 10m x 5m x 4m has a mass of 80 000 kg. What is the density of this wood?
A. 2000kg/m^{2} B. 4000kg/m^{2 }C. 200 kg/m^{2} D. 400 kg/m^{2}
3. The density of gold is 19.3 g/cm^{3}. What is the mass of 10cm^{3} gold?
A. 19.3 g B. 0.193g C. 1.93g D. 193g
4.. What is the mass of the copper cube having each side 2cm? ( take c_{opper} = 9g/cm^{3})
A. 0.18g B. 72g C. 180g D. 36g
5. What is the volume of 60g wood ? ( d_{ wood} = 0. 6 g / cm^{3} )
A. 10cm^{3 } B. 36cm^{3 }C. 100cm^{3 }D. 360cm^{3}
6. Study the table below and use it to spot the correct answer.
Material | Density (g/cm^{3}) | Mass (g_) |
K | 3 9 6 5 | 60 180 360 200 |
From the values shown in the table which material has the biggest volume?
A. K B. L C. M D. N
7. What is the volume and mass of the block which measures by 2m, by 3m by 5m if its density is 1500kg/m^{3}?
A. 50m^{3} 75 000 kg B. 100 m^{3} 75 000 kg
C. 30m^{3} 75 000 kg D. 30 m^{3} 75 000 kg
8. Two litres of corn oil has a mass of 1. 85kg. What is the density of the oil?
A. 1850kg/m^{2} B. 925kg/m^{2 }C. 185kg/m^{2} D. 92.5kg/m^{2}
9. If an object of volume 0.02m^{3} weighs 500 N in a liquid of density 2000kg/m^{3}, what is the weight in air?
A. 900 N B. 1000 N C. 400 N D. 600 N
10. Which one of the following is the SI unit of density?
A. kgm^{3} B. kg/m^{-3} C. g/cm^{3} D. kg/m^{3})
11. If 10g water and 10cm^{3} alcohol are mixed what will be the mass of the mixture?
( d _{alcohol }= 0.80 g/cm^{3} )
A. 18g B. 20g C. 16g D. 19g
12. A tin containing 5 litres of paint has a mass of 8.5kg. The mass of the empty tin is 2.0kg, the density of the paint is
A. 1.3kg m^{-3} B. 1.3×10^{3}kg m^{-3} C. 1.7×10^{3}kg m^{-3} D. 2.1×10^{3}kg m^{-3}
13. A rectangular block of tin is 0.5m long and 0.01m thick. Find the width of the block if its mass and density are 0.45kg and 9000 kg m^{-3} respectively.
A. B.
C. D.
14. A box of dimensions 0.2m by 0.3m by 0.5m is full of a gas of density 200kg/m^{3}. The mass of the gas is
A. 3×10^{-2}kg B. 6.0×10^{0}kg C. 2×10^{2}kg D. 6.7×10^{3}kg
15. A piece of material of mass 200g has a density of 25kgm^{-3}. Calculate its volume in m^{3}.
A. B. C. D.
16. Two solid cubes have the same mass but their edges are in the ratio 4:1. What is the ratio of their densities?
A. 1:4 B. 1:8 C. 1:16 D. 1: 64
17. A tin containing 6×10^{-3} m^{3} of paint has a mass of 8kg. If the mass of the empty tin with the lid is 0.5kg, calculate the density of the paint in kgm^{-3}
A. B. C. D.
18 A tank 2 m tall base area of 2.5 m^{2} is filled to the brim with a liquid which exerts a force of 40000N at the bottom. Calculate the density of the liquid.
A. B. C. D.
19. The following reading were recorded when measuring the density of a stone; Mass of the stone = 25g, volume of water = 25 cm^{3}, volume of water and
stone = 35cm^{3} .What is the density of the stone?
A. B. C. D.
20. Liquid Y of a volume 0.40m^{3} and density 900 kg/m^{3} is mixed with liquid Z of volume 0.35 m^{3} and density 800 kg/m^{3}. Calculate the density of the mixture.
A. 800kg/m^{3} B. 840 kg/m^{3} C. 850 kg/m^{3} D. 900 kg/m^{3}