In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of

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Q. In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage carefully and answer the following questions:

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of lycopodium powder?

(d) How will you calculate the volume of

n

drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

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Concept understanding — Atomic Volume Calculation

Atomic Volume: Meaning and Calculation

Atoms are mostly empty space — a tiny dense nucleus wrapped in a fuzzy electron cloud. So "atomic volume" does not mean the volume of a solid ball; it means the average space one atom occupies when many atoms are packed together in a solid or liquid.

Think of a crowded hall: to find the space per person you divide the hall's volume by the number of people. Atomic volume does exactly that for atoms.

Definition

Atomic volume is the volume occupied by one mole of atoms of an element in its solid or liquid state. It is found from the element's molar mass and density:

V_{\text{atomic}} = \frac{M}{\rho}

$V_{\text{atomic}}$ = atomic volume (cm³/mol)
$M$ = molar mass (g/mol)
$\rho$ = density (g/cm³)

This gives the volume per mole. Dividing by Avogadro's number gives the space per single atom:

V_{\text{one atom}} = \frac{V_{\text{atomic}}}{N_A}, \qquad N_A = 6.022\times10^{23}\ \text{mol}^{-1}

Worked example — aluminium

Molar mass $M = 26.98$ g/mol, density $\rho = 2.70$ g/cm³:

V_{\text{atomic}} = \frac{26.98}{2.70} = 9.99\ \text{cm}^3/\text{mol}

So one mole of Al atoms occupies about 10 cm³. Per atom:

V_{\text{one Al atom}} = \frac{9.99}{6.022\times10^{23}} = 1.66\times10^{-23}\ \text{cm}^3

Periodic trends

Down a group: atomic volume increases — more electron shells make atoms larger.
Across a period: it generally decreases — rising nuclear charge pulls the electrons in tighter.
Allotropes differ: diamond is denser than graphite, so diamond's atomic volume is smaller, though both are carbon.

These trends help explain why alkali metals (large atomic volume) are soft and reactive, while transition metals (smaller atomic volume) are hard and dense. …

Why this formula?

Atomic Volume Calculation: Understanding the "Why" Behind the Formula

What Is Atomic Volume?

Atomic volume is not the volume of a single atom — it's the volume occupied by one mole of atoms of an element in its solid state. This is a macroscopic quantity that helps us understand how tightly atoms pack together.

The key formula is:

\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}

Let's break down why this works.

The Core Reasoning: From Mass to Volume

Step 1: What does density tell us?

Density ( $\rho$ ) is defined as:

\rho = \frac{\text{Mass}}{\text{Volume}}

For a pure solid element, if we take one mole of atoms:

Mass of one mole = Atomic mass (in g/mol)
Volume of one mole = Atomic volume (in cm³/mol)

So:

\rho = \frac{\text{Atomic mass}}{\text{Atomic volume}}

Step 2: Rearranging to find atomic volume

\text{Volume} = \frac{\text{Mass}}{\text{Density}}

Therefore:

\boxed{\text{Atomic Volume} = \frac{\text{Atomic Mass}}{\text{Density}}}

Why This Makes Physical Sense

Atomic mass tells you how heavy one mole of atoms is; density tells you how much mass fits in a given space. Dividing mass by density gives the space that mass occupies.

Example intuition: If iron has atomic mass ≈ 56 g/mol and density 7.87 g/cm³, then:

\text{Atomic volume} = \frac{56}{7.87} \approx 7.1 \text{ cm}^3/\text{mol}

This means one mole of iron atoms (about $6.022 \times 10^{23}$ atoms) occupies roughly 7.1 cm³ of space.

Important Exam Points

Concept	Why It Matters
Units	Atomic mass in g/mol, density in g/cm³ → atomic volume in cm³/mol
Solid state only	The formula assumes atoms are closely packed; gases/liquids have different packing

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The experiment uses serial dilution to create a very dilute oleic acid solution, allowing a single drop to form a measurable monolayer on water; the final solution contains $\frac{1}{400}$ mL of oleic acid per mL of solution.

The experiment described is a classic method to estimate the size of a molecule, specifically the length of an oleic acid molecule. The core idea is to create a very thin, single-molecule layer (a monolayer) of oleic acid on a water surface. By knowing the volume of oleic acid in this layer and measuring the area it covers, we can calculate the thickness of the layer, which corresponds to the length of a single oleic acid molecule.

To achieve a monolayer, the oleic acid must be extremely dilute. This is why a process of serial dilution is employed. Oleic acid molecules are long-chain fatty acids with a polar (hydrophilic) head and a non-polar (hydrophobic) tail. When placed on water, the polar head interacts with water, while the non-polar tail points away from it, forming a stable monolayer.

Let's address each part of the question.

Why do we dissolve oleic acid in alcohol?

Oleic acid itself is largely insoluble in water. If a drop of pure oleic acid were placed on water, it would likely form a thick, irregular blob rather than spreading into a uniform, single-molecule layer. Alcohol, on the other hand, is a good solvent for oleic acid. When the oleic acid-alcohol solution is dropped onto water:
- The alcohol helps the oleic acid spread out evenly over the water surface.
- Alcohol is volatile, meaning it evaporates quickly, leaving behind only the oleic acid molecules.
- This process facilitates the formation of a stable, uniform monolayer of oleic acid, with the polar heads interacting with the water and the non-polar tails pointing upwards.
What is the role of lycopodium powder?

The oleic acid film formed on the water surface is extremely thin and transparent, making its boundaries difficult to see with the naked eye. Lycopodium powder consists of fine, hydrophobic spores. When sprinkled evenly over the water surface before adding the oleic acid solution:
- The oleic acid, as it spreads, pushes the lycopodium powder away from the area it occupies.
- This creates a clear, visible boundary between the area covered by the oleic acid film (devoid of powder) and the surrounding water surface (covered with powder).
- This visible boundary allows for accurate measurement of the diameter (and thus the area) of the oleic acid film.
What would be the volume of oleic acid in each mL of solution prepared?

This involves a two-step dilution process.
- Step 1: First Dilution 1 mL of oleic acid is dissolved in 19 mL of alcohol. The total volume of the first solution is $1 \text{ mL} + 19 \text{ mL} = 20 \text{ mL}$ . The concentration of oleic acid in this first solution is:

C_1 = \frac{\text{Volume of oleic acid}}{\text{Total volume of solution}} = \frac{1 \text{ mL}}{20 \text{ mL}} = \frac{1}{20}

    So, each mL of this first solution contains $\frac{1}{20}$ mL of oleic acid.

*   **Step 2: Second Dilution**
    1 mL of the *first* solution is taken and diluted to 20 mL by adding alcohol.
    The volume of oleic acid in this 1 mL of the first solution is $\frac{1}{20}$ mL.
    This $\frac{1}{20}$ mL of oleic acid is now present in a total volume of 20 mL (the final solution).
    The concentration of oleic acid in the final solution is:

C_2 = \frac{\text{Volume of oleic acid}}{\text{Total volume of final solution}} = \frac{\frac{1}{20} \text{ mL}}{20 \text{ mL}} = \frac{1}{20 \times 20} = \frac{1}{400}

    Therefore, each mL of the final diluted solution contains $\frac{1}{400}$ mL of oleic acid.

4. How will you calculate the volume of $n$ drops of this solution of oleic acid?

Measuring the volume of a single drop directly is difficult and prone to significant error due to its small size and variations in drop formation. A more accurate method is to measure the total volume of a larger number of drops.

*   **Procedure:**
    1.  Take a burette or a graduated pipette.
    2.  Carefully count $n$ drops of the final oleic acid solution as they fall into a measuring cylinder or another graduated container.
    3.  Record the total volume of these $n$ drops, let's call it $V_n$.
    4.  The volume of $n$ drops is simply $V_n$. …

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Concept: Molecular Size Estimation by Monolayer Film Method (Oleic Acid Experiment)

This is a standard thin film method to estimate the size of a molecule. The key idea: if a known volume of a substance spreads into a monolayer (one molecule thick) on water, then:

\text{Thickness of film} = \frac{\text{Volume of substance}}{\text{Area of film}} = \text{Size of one molecule}

(a) Why do we dissolve oleic acid in alcohol?

Method: Solubility and spreading principle.

Steps:

Oleic acid alone does not spread uniformly on water — it forms droplets.
Alcohol acts as a solvent that:
- Dissolves oleic acid completely.
- Evaporates quickly after the drop is placed on water.
- Leaves behind a thin, uniform layer of oleic acid molecules.

Answer: Alcohol helps in uniform spreading of oleic acid and then evaporates, leaving only the oleic acid film.

(b) What is the role of lycopodium powder?

Method: Visual boundary detection.

Steps:

Before placing the drop, sprinkle lycopodium powder evenly on the water surface.
When the oleic acid solution spreads, it pushes the powder aside, creating a clear circular boundary.
This allows accurate measurement of the diameter (and hence area) of the film.

Answer: Lycopodium powder makes the edge of the film visible so the area can be measured.

(c) Volume of oleic acid in each mL of the final solution?

Method: Stepwise dilution calculation.

Steps:

First dilution: 1 mL oleic acid + 19 mL alcohol → total 20 mL solution.
- Volume of oleic acid per mL = $\frac{1}{20}$ mL
Second dilution: 1 mL of this solution is diluted to 20 mL.
- Volume of oleic acid in this 1 mL taken = $\frac{1}{20}$ mL
- After diluting to 20 mL, concentration becomes:

\frac{1/20}{20} = \frac{1}{400} \text{ mL per mL of final solution}

Answer: $\boxed{\frac{1}{400} \text{ mL}}$ of oleic acid per mL of final solution.

(d) How to calculate the volume of $n$ drops?

Method: Drop volume by counting.

Steps:

Take a graduated burette or pipette filled with the solution.
Count the number of drops ( $N$ ) that make a known volume (say 1 mL). …

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Here are the common mistakes students make on this classic experimental determination of molecular size problem, along with how to avoid each.

(a) Why do we dissolve oleic acid in alcohol?

Common Mistake:

Saying "to make a solution" without explaining why alcohol is specifically chosen.
Confusing the role of alcohol with water (e.g., "to dissolve it in water").

Why it’s wrong:

Oleic acid does not dissolve in water. If you tried to put pure oleic acid on water, it would form a thick lump, not a monomolecular layer. Alcohol acts as a volatile solvent — it dissolves the oleic acid and then evaporates quickly once the drop hits the water surface, leaving only the oleic acid to spread.

How to avoid:

Remember the key property: Oleic acid is soluble in alcohol but not in water. Alcohol is a carrier that evaporates.
Write the logic: "Alcohol dissolves oleic acid and then evaporates, leaving a thin film of pure oleic acid on the water surface."

(b) What is the role of lycopodium powder?

Common Mistake:

Saying "to make the film visible" or "to colour the film."
Thinking it helps the film spread.

Why it’s wrong:

Lycopodium powder is hydrophobic (water-repelling). It floats on the water surface. When the oleic acid film spreads, it pushes the powder aside, creating a clear, circular boundary. This allows you to measure the diameter of the film accurately.

How to avoid:

Key phrase: "To mark the boundary of the film."
Visualise: The powder is like a fence — the film pushes it away, so you can see exactly where the film ends.

(c) What would be the volume of oleic acid in each mL of solution prepared?

Common Mistake:

Forgetting the two-step dilution.
Directly using the initial 1 mL in 20 mL ratio without accounting for the second dilution.

Why it’s wrong:

The problem describes two dilutions:

1 mL oleic acid + 19 mL alcohol → total 20 mL solution. Concentration = $\frac{1}{20}$ mL oleic acid per mL of solution.
Then 1 mL of this solution is diluted to 20 mL with alcohol. So the final concentration = $\frac{1}{20} \times \frac{1}{20} = \frac{1}{400}$ mL oleic acid per mL of final solution.

How to avoid:

Write the dilution steps clearly:
- Step 1: $C_1 = \frac{1}{20}$
- Step 2: $C_2 = C_1 \times \frac{1}{20} = \frac{1}{400}$
Answer: $\boxed{\frac{1}{400} \text{ mL}}$

(d) How will you calculate the volume of $n$ drops of this solution of oleic acid?

Common Mistake:

Assuming you know the volume of one drop without mentioning how to find it.
Writing a formula like $V = n \times \text{(volume of one drop)}$ but not explaining how to get the volume of one drop.

Why it’s wrong:

You cannot assume the drop volume. In the lab, you measure it using a burette or dropper:

Count the number of drops ( $N$ ) that make up a known volume (e.g., 1 mL).
Then volume of one drop = $\frac{1 \text{ mL}}{N}$ .
For $n$ drops: $V = n \times \frac{1}{N}$ mL.

How to avoid:

Always mention the experimental step: "First, find the volume of one drop by counting drops in 1 mL."
Formula: $V_{n\text{ drops}} = n \times \frac{1}{N}$ where $N$ is the number of drops per mL.

(e) What will be the volume of oleic acid in one drop of this solution?

Common Mistake: …

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