Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes

Q: Why do we cube the radius when finding the volume?

The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$. Since the drops are spherical, we must cube the given radius to find the space they occupy.

Q: Why is the volume of the film equal to the volume of the drops?

This is based on the principle of **conservation of volume**. When the drops spread out to form the film, the total amount of liquid doesn't change, only its shape does.

Q: Why do we multiply the film thickness by 0.01 at the end?

The film we formed is made of the *solution*, not pure oleic acid. The concentration tells us that only $0.01 \text{ cm}^3$ of every $1 \text{ cm}^3$ of solution is oleic acid. So, the oleic acid layer is only $1\%$ of the total thickness.

Q: How do we handle the unit conversions from cm to m?

We calculated the thickness in centimeters. To convert to meters, we multiply by $10^{-2}$. So, $25 \times 10^{-10} \text{ cm}$ becomes $25 \times 10^{-12} \text{ m}$.

Q: What does 'monomolecular thickness' mean in this context?

It means the oleic acid spreads out so thin that it forms a layer exactly one molecule thick. This is a classic experimental technique used to estimate the size of a single molecule!

Question

Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes $0.01 	ext{ cm}^3$ of oleic acid per $	ext{cm}^3$ of the solution. Then, you make a thin film of this solution (monomolecular thickness) of area $4 	ext{ cm}^2$ by considering 100 spherical drops of radius $\left(\frac{3}{40\pi}ight)^{1/3} 	imes 10^{-3} 	ext{ cm}$. Then, the thickness of oleic acid layer will be $x 	imes 10^{-14} 	ext{ m}$, where $x$ is ……………… .

Sigma Percentile · Accepted Answer

### Measuring the Invisible: The Oleic Acid Experiment

Imagine trying to measure the size of a single molecule. It sounds like a task for a multi-million dollar electron microscope, right? But what if I told you that you could estimate it using just a few drops of liquid and some basic geometry? This is the beauty of the classic oleic acid experiment.

In this problem, we are given 100 tiny spherical drops of a dilute oleic acid solution. When these drops are placed on a surface, they spread out to form a very thin, flat film. Our ultimate goal is to find the thickness of just the oleic acid molecules within this film.

### The Setup: Drops to Film

First, we need to find the total volume of the liquid we are working with. We have 100 drops, and we know the formula for the volume of a single sphere is $V = \frac{4}{3}\pi r^3$. Therefore, the total volume for 100 drops is simply:

$$V = 100 	imes \frac{4}{3}\pi r^3$$

We are given a rather intimidating expression for the radius of each drop: $r = \left(\frac{3}{40\pi}ight)^{1/3} 	imes 10^{-3} 	ext{ cm}$. But don't let that scare you! Notice the power of $1/3$ in the radius. When we substitute this into our volume equation and cube it, that fractional power is going to cancel out beautifully.

### Calculating the Volume

Let's carefully substitute the radius and do the math:

$$V = 100 	imes \frac{4}{3}\pi \left( \left(\frac{3}{40\pi}ight)^{1/3} 	imes 10^{-3} ight)^3$$

Cubing the radius removes the $1/3$ power, leaving $\frac{3}{40\pi}$, and $10^{-3}$ becomes $10^{-9}$. The $\pi$ cancels out perfectly, and after simplifying the fractions, we get:

$$V = 100 	imes \frac{4}{3}\pi 	imes \frac{3}{40\pi} 	imes 10^{-9} = 10^{-8} 	ext{ cm}^3$$

This is the total volume of the 100 drops.

### The Film Thickness

Now, these 100 drops spread out to form a thin film with an area of $4 	ext{ cm}^2$. By the principle of conservation of volume, the volume of this film must be equal to the total volume of the drops we just calculated. The volume of a cylindrical film is simply its area multiplied by its thickness, $t$:

$$V_{	ext{film}} = A 	imes t$$

Equating the two volumes, we have:

$$4 	imes t = 10^{-8}$$

Solving for $t$ gives us the thickness of the solution film:

$$t = 0.25 	imes 10^{-8} 	ext{ cm} = 25 	imes 10^{-10} 	ext{ cm}$$

Converting this to meters (by multiplying by $10^{-2}$), we get $t = 25 	imes 10^{-12} 	ext{ m}$.

### The Final Layer: Concentration Matters

But wait, there is a catch! The thickness we just found is for the *entire* solution film. The question specifically asks for the thickness of the *oleic acid layer*.

We are told the concentration is $0.01 	ext{ cm}^3$ of oleic acid per $	ext{cm}^3$ of solution. This means the oleic acid makes up only $1\%$ of the total film thickness. To find the thickness of the pure oleic acid layer ($t_0$), we multiply our film thickness by this concentration:

$$t_0 = 0.01 	imes 25 	imes 10^{-12} = 25 	imes 10^{-14} 	ext{ m}$$

Comparing this result with the given expression $x 	imes 10^{-14} 	ext{ m}$, we can clearly see that **$x = 25$**. And that is our final answer! A brilliant application of simple geometry to measure the microscopic world.

Enter Numerical Value:

Visualized Solution

\text{The Oleic Acid Experiment}

\text{Volume of } 100 \text{ Drops}

\text{Substituting the Radius}

\text{Calculating Total Volume}

\text{Volume of the Film}

\text{Thickness of Solution Film}

\text{Thickness of Oleic Acid Layer}

\text{Final Calculation}

The Sigma Insight: Dimensional Analysis

Measuring the Invisible

The Setup

Calculating the Volume

The Film Thickness

The Final Layer

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