1. Overview
Density is a fundamental property of matter that describes how much mass is concentrated in a specific volume. Understanding density allows us to predict whether objects will sink or float and helps in identifying unknown substances based on their physical characteristics.
Key Definitions
- Density: The mass per unit volume of a substance.
- Mass: The amount of matter in an object, measured in grams (g) or kilograms (kg).
- Volume: The amount of space an object takes up, measured in cubic centimeters ($\text{cm}^3$) or cubic meters ($\text{m}^3$).
- Displacement: The volume of fluid pushed out of the way when an object is submerged.
Core Content
The Density Formula
Density is calculated by dividing the mass of an object by its volume. $$\rho = \frac{m}{V}$$ (Note: The symbol for density is the Greek letter rho, $\rho$, which looks like a curly 'p').
Determining Density: Experimental Methods
1. Regularly Shaped Solid (e.g., a cube or cuboid)
- Measure the mass ($m$) using a digital balance.
- Measure the length, width, and height using a ruler.
- Calculate volume ($V = l \times w \times h$).
- Apply the formula $\rho = m/V$.
2. A Liquid
- Place an empty measuring cylinder on a digital balance and "tare" (zero) it, or record the mass of the empty cylinder.
- Pour the liquid into the cylinder and record the volume ($V$) from the scale.
- Record the new mass and subtract the mass of the empty cylinder to find the mass of the liquid ($m$).
- Apply the formula $\rho = m/V$.
3. Irregularly Shaped Solid (Displacement Method)
- Measure the mass ($m$) of the object using a balance.
- Fill a measuring cylinder with a known volume of water ($V_1$).
- Carefully submerge the object in the water.
- Record the new volume ($V_2$).
- Calculate the volume of the object: $V = V_2 - V_1$.
- Apply the formula $\rho = m/V$.
Floating and Sinking
- An object will float if its density is less than the density of the liquid.
- An object will sink if its density is greater than the density of the liquid.
- Example: Water has a density of $1.0 \text{ g/cm}^3$. A piece of wood with a density of $0.7 \text{ g/cm}^3$ will float, while a stone with a density of $2.5 \text{ g/cm}^3$ will sink.
Extended Content (Extended Only)
Floating Liquids (Immiscible Liquids)
When two liquids that do not mix (immiscible) are poured into the same container, they will form layers based on their densities.
- The liquid with the lowest density will float on the top.
- The liquid with the highest density will sink to the bottom.
Worked Example: A beaker contains Liquid A ($\rho = 0.8 \text{ g/cm}^3$) and Liquid B ($\rho = 1.2 \text{ g/cm}^3$). If they are immiscible, which liquid is on top?
- Answer: Liquid A will float on Liquid B because $0.8 \text{ g/cm}^3 < 1.2 \text{ g/cm}^3$.
Key Equations
| Equation | Symbols | Units |
|---|---|---|
| $\rho = \frac{m}{V}$ | $\rho$ = density, $m$ = mass, $V$ = volume | $\text{g/cm}^3$ or $\text{kg/m}^3$ |
| $V = l \times w \times h$ | $l$ = length, $w$ = width, $h$ = height | $\text{cm}^3$ or $\text{m}^3$ |
| $V_{obj} = V_2 - V_1$ | $V_2$ = final volume, $V_1$ = initial volume | $\text{cm}^3$ or $\text{ml}$ |
Unit Conversion Tip: To convert from $\text{g/cm}^3$ to $\text{kg/m}^3$, multiply by 1,000. (e.g., $1 \text{ g/cm}^3 = 1,000 \text{ kg/m}^3$)
Common Mistakes to Avoid
- ❌ Wrong: Inverting the formula (calculating Volume ÷ Mass).
- ✅ Right: Always divide Mass by Volume.
- ❌ Wrong: Forgetting to subtract the mass of the beaker when measuring a liquid.
- ✅ Right: Liquid mass = (Mass of beaker + liquid) - (Mass of empty beaker).
- ❌ Wrong: Using the total final volume ($V_2$) as the object's volume in displacement.
- ✅ Right: Subtract the initial water level from the final level ($V_2 - V_1$).
- ❌ Wrong: Miscalculating volume of a cube by only multiplying two sides.
- ✅ Right: For a 3D shape, you must multiply length × width × height.
Exam Tips
- Read the Meniscus: When measuring volume in a cylinder, always read from the bottom of the curve (the meniscus) at eye level to avoid parallax error.
- Check the Units: If the mass is in kg and volume is in $\text{m}^3$, the density must be $\text{kg/m}^3$. Do not mix grams and cubic meters in the same calculation.
- Show Your Working: Even if your final answer is wrong, you can earn marks for correctly stating the formula ($\rho = m/V$) and showing the subtraction for displacement or mass.
Exam-Style Questions
Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.
Exam-Style Question 1 — Short Answer [5 marks]
Question:
A student is given a small, irregularly shaped rock and is asked to determine its density.
(a) State the formula used to calculate density. [1]
(b) Describe how the student can accurately determine the volume of the rock using the displacement method and standard laboratory equipment. [4]
Worked Solution:
(a)
- Density is mass divided by volume. $density = \frac{mass}{volume}$ This is the definition of density.
How to earn full marks:
- State the formula correctly, either in words or symbols.
(b)
Fill a measuring cylinder with a known volume of water (e.g., 50 cm$^3$). This establishes the initial volume.
Record the initial volume of water in the measuring cylinder. Write down the starting point.
Carefully lower the rock into the measuring cylinder, ensuring it is fully submerged. The rock displaces its own volume of water.
Record the new volume of water in the measuring cylinder. Write down the final volume.
Calculate the volume of the rock by subtracting the initial volume of water from the final volume of water. The volume of the rock = final volume - initial volume. The difference in volumes is the rock's volume.
How to earn full marks:
- Mention using a measuring cylinder and water.
- State to record the initial volume.
- State to carefully lower the rock into the water.
- State to subtract the initial volume from the final volume to find the rock's volume.
Common Pitfall: Remember that the displacement method relies on the object being fully submerged. Also, make sure you're subtracting the volumes in the correct order (final - initial) to get a positive volume for the object.
Exam-Style Question 2 — Short Answer [6 marks]
Question:
Two metal blocks, A and B, are made from different materials. Block A has a volume of 2.0 x 10$^{-5}$ m$^3$ and a mass of 0.16 kg. Block B has a volume of 5.0 x 10$^{-5}$ m$^3$ and a mass of 0.35 kg.
(a) Calculate the density of block A. [2]
(b) Calculate the density of block B. [2]
(c) State which block, A or B, is made from a denser material. [1]
(d) Explain how you can tell which block is denser based on your calculations. [1]
Worked Solution:
(a)
- Calculate the density of Block A using the formula density = mass / volume. $density = \frac{0.16 , kg}{2.0 \times 10^{-5} , m^3} = 8000 , kg/m^3$ Substituting the known values into the density equation.
How to earn full marks:
- Correct substitution of mass and volume for block A.
- Correct calculation and unit. $\boxed{8000 , kg/m^3}$
(b)
- Calculate the density of Block B using the formula density = mass / volume. $density = \frac{0.35 , kg}{5.0 \times 10^{-5} , m^3} = 7000 , kg/m^3$ Substituting the known values into the density equation.
How to earn full marks:
- Correct substitution of mass and volume for block B.
- Correct calculation and unit. $\boxed{7000 , kg/m^3}$
(c)
- Block A is made from the denser material. Direct comparison of the calculated densities.
How to earn full marks:
- State block A.
(d)
- Block A has a higher calculated density value than Block B. Linking the density value to the material.
How to earn full marks:
- State that Block A has a higher density value.
Common Pitfall: Make sure you're using the correct units (kg for mass and m$^3$ for volume) when calculating density. If you're given grams and cm$^3$, you'll need to convert them to kg and m$^3$ before using the density formula, or calculate in g/cm$^3$ and convert the final answer.
Exam-Style Question 3 — Extended Response [9 marks]
Question:
A student investigates whether a small stone will float or sink in different liquids. The student measures the mass of the stone to be 45 g and determines its volume to be 15 cm$^3$.
(a) Calculate the density of the stone in g/cm$^3$. [2]
(b) The student has two liquids: Liquid X with a density of 0.8 g/cm$^3$ and Liquid Y with a density of 3.2 g/cm$^3$. Predict whether the stone will float or sink in each liquid. Explain your reasoning. [4]
(c) The student then repeats the experiment with a piece of wood. The wood has a volume of 20 cm$^3$. It floats in Liquid X with 75% of its volume submerged. Calculate the mass of the piece of wood. [3]
Worked Solution:
(a)
- Calculate the density of the stone using the formula density = mass / volume. $density = \frac{45 , g}{15 , cm^3} = 3 , g/cm^3$ Substituting the known values into the density equation.
How to earn full marks:
- Correct substitution of mass and volume for the stone.
- Correct calculation and unit. $\boxed{3 , g/cm^3}$
(b)
The stone will sink in Liquid X because the density of the stone (3 g/cm$^3$) is greater than the density of Liquid X (0.8 g/cm$^3$). Comparing the densities to determine sinking.
The stone will sink in Liquid Y because the density of the stone (3 g/cm$^3$) is less than the density of Liquid Y (3.2 g/cm$^3$). Comparing the densities to determine floating.
How to earn full marks:
- State that the stone sinks in Liquid X.
- Explain that the stone's density is greater than Liquid X's density.
- State that the stone sinks in Liquid Y.
- Explain that the stone's density is less than Liquid Y's density.
(c)
Calculate the volume of the wood submerged in Liquid X: 75% of 20 cm$^3$ = 0.75 x 20 cm$^3$ = 15 cm$^3$. Find the volume of the displaced liquid.
Since the wood is floating, the weight of the wood equals the weight of the liquid displaced. Therefore, the mass of the wood equals the mass of the displaced liquid. Using Archimedes' principle.
Calculate the mass of the displaced liquid (and thus the mass of the wood): mass = density x volume = 0.8 g/cm$^3$ x 15 cm$^3$ = 12 g. Applying the density formula to find the mass.
How to earn full marks:
- Calculate the submerged volume: 15 cm$^3$.
- State that the weight of the wood equals the weight of the displaced liquid (or equivalent).
- Correct calculation and unit. $\boxed{12 , g}$
Common Pitfall: In part (b), remember that an object floats if its density is less than the density of the liquid. Also, in part (c), the percentage submerged tells you the volume of liquid displaced, which is key to finding the mass of the wood.
Exam-Style Question 4 — Extended Response [10 marks]
Question:
A student wants to determine the density of a small, irregularly shaped metal key. The student has access to a balance, a measuring cylinder, water, and a thin thread.
(a) Describe a detailed procedure the student should follow to accurately determine the density of the key. Your procedure should include all necessary measurements and calculations. [6]
(b) The student records the following data: - Mass of the key: 24.0 g - Initial volume of water in the measuring cylinder: 50.0 cm$^3$ - Final volume of water in the measuring cylinder after the key is submerged: 53.0 cm$^3$ Calculate the density of the key in kg/m$^3$. [3]
(c) Suggest one source of error in this experiment and explain how this error would affect the calculated density of the key. [1]
Worked Solution:
(a)
Use the balance to measure the mass of the key. Record the mass, $m$, in grams (g). Measure the key's mass accurately.
Fill the measuring cylinder with a known volume of water (e.g., 50.0 cm$^3$). Record this initial volume, $V_1$, in cm$^3$. Establish the starting water level.
Carefully tie the key to the thin thread and gently lower the key into the measuring cylinder until it is fully submerged. Ensure no water splashes out. Submerge the key without losing water.
Record the new volume of water in the measuring cylinder, $V_2$, in cm$^3$. Read the water level after submersion.
Calculate the volume of the key by subtracting the initial volume of water from the final volume of water: $V_{key} = V_2 - V_1$. The volume will be in cm$^3$. Find the key's volume by displacement.
Calculate the density of the key using the formula: $density = \frac{mass}{volume} = \frac{m}{V_{key}}$. State the density with appropriate units, g/cm$^3$. To convert to kg/m$^3$, multiply the result by 1000. Calculate density from mass and volume.
How to earn full marks:
- State to measure the mass of the key using a balance.
- State to record the initial volume of water.
- State to carefully lower the key into the water using thread.
- State to record the final volume of water.
- State to calculate the volume of the key by subtracting the initial volume from the final volume.
- State to calculate the density using the formula density = mass/volume and mention the unit g/cm$^3$.
(b)
Calculate the volume of the key: $V_{key} = V_2 - V_1 = 53.0 , cm^3 - 50.0 , cm^3 = 3.0 , cm^3$. Find the volume by subtraction.
Calculate the density of the key in g/cm$^3$: $density = \frac{24.0 , g}{3.0 , cm^3} = 8.0 , g/cm^3$. Apply the density formula.
Convert the density to kg/m$^3$: $8.0 , g/cm^3 \times 1000 = 8000 , kg/m^3$. Convert units correctly.
How to earn full marks:
- Correct calculation of the volume of the key: 3.0 cm$^3$.
- Correct calculation of the density in g/cm$^3$: 8.0 g/cm$^3$.
- Correct conversion to kg/m$^3$ with correct unit. $\boxed{8000 , kg/m^3}$
(c)
Possible error: Some water may splash out of the measuring cylinder when the key is submerged. This would result in a smaller measured volume for the key. Identify a plausible experimental error.
Effect: A smaller calculated volume would lead to a higher calculated density for the key, as density is inversely proportional to volume. Explain the effect of the error on the density.
How to earn full marks:
- State a plausible source of error (e.g., water splashing out).
- Explain that this error would lead to a higher calculated density.
Common Pitfall: Be very careful when converting between g/cm$^3$ and kg/m$^3$. Multiplying by 1000 is the correct conversion from g/cm$^3$ to kg/m$^3$. Also, remember that if water splashes out, the volume measurement will be lower than it should be, leading to an overestimation of the density.