0.27 g of a long chain fatty acid was dissolved in 100 cm^3 of hexane. 10 mL of this solution was ad

Q: Why do we calculate the mass for 10 mL instead of using the 0.27 g directly?

The 0.27 g is dissolved in 100 mL of hexane. However, only 10 mL of this solution is actually used to form the monolayer. Therefore, we must find the mass of the fatty acid in just that 10 mL portion.

Q: What happens to the hexane? Why don't we include its volume?

The problem states that the 'hexane evaporates'. Hexane is a highly volatile solvent used just to spread the fatty acid. Once it evaporates, only the pure fatty acid remains to form the monolayer.

Q: Why do we assume the monolayer is a cylinder?

A watch glass is circular, and the drop spreads out uniformly from the center to the edge, forming a circular shape with a very small, uniform thickness (height). Geometrically, a circle with a uniform thickness is a cylinder.

Q: Why is the value of $\pi$ taken as 3 instead of 3.14?

The problem explicitly provides the approximation $[\pi = 3]$ in the given data. Always use the approximations provided in the question to ensure your answer matches the options exactly and to simplify calculations.

Q: I got $10^{-4}$, why is the answer $10^{-6}$?

You likely found the height in centimeters ($10^{-4} \text{ cm}$). The options are given in meters. You must convert centimeters to meters by multiplying by $10^{-2}$, which gives $10^{-6} \text{ m}$.

Question

0.27 g of a long chain fatty acid was dissolved in $100 	ext{ cm}^3$ of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer?
[Density of fatty acid $= 0.9 	ext{ g cm}^{-3}$; $\pi = 3$]

Options:

(A) $10^{-6} 	ext{ m}$

(B) $10^{-4} 	ext{ m}$

(C) $10^{-8} 	ext{ m}$

(D) $10^{-2} 	ext{ m}$

Sigma Percentile · Accepted Answer

This problem is a beautiful intersection of basic stoichiometry, physical properties, and simple geometry. It tests your ability to track a substance through a dilution and then apply its physical characteristics to a real-world geometric model. Let's break down the journey of this fatty acid drop by drop.

### The Setup: From Solution to Surface

Imagine you have a stock solution: $100 	ext{ cm}^3$ (which is the same as $100 	ext{ mL}$) of hexane containing exactly $0.27 	ext{ g}$ of a long-chain fatty acid.

We don't pour the whole thing out. We carefully extract just a $10 	ext{ mL}$ portion of this solution and drop it onto the surface of water in a watch glass. Hexane is a highly volatile solvent, meaning it evaporates very quickly into the air. Once the hexane is gone, what is left behind? Only the pure fatty acid, which spreads out over the water to form a very thin, circular layer known as a **monolayer**.

### The Mass and Volume Connection

Before we can figure out the dimensions of this monolayer, we need to know exactly how much fatty acid we are dealing with. Since the solution is uniform, we can use a simple unitary method to find the mass in our $10 	ext{ mL}$ sample.

If $100 	ext{ mL}$ contains $0.27 	ext{ g}$, then $10 	ext{ mL}$ will contain one-tenth of that mass:
$$m = \frac{0.27}{100} 	imes 10 = 0.027 	ext{ g}$$

Now we have the mass of the fatty acid forming the monolayer. But geometry deals with space, not mass. We need to convert this mass into a volume. The problem provides the crucial bridge: the density of the fatty acid is $d = 0.9 	ext{ g cm}^{-3}$.

Using the classic relationship $V = \frac{m}{d}$, we can find the volume occupied by the fatty acid:
$$V = \frac{0.027 	ext{ g}}{0.9 	ext{ g cm}^{-3}} = 0.03 	ext{ cm}^3$$

### The Geometry of the Monolayer

The fatty acid spreads out in a circle on the watch glass. The problem states the distance from the edge to the center is $10 	ext{ cm}$. This is simply the radius ($r$) of our circular layer.

Because the layer has some thickness (even if it's microscopic), it forms a very flat cylinder. The volume of a cylinder is given by the formula:
$$V = \pi r^2 h$$

We know the volume ($V = 0.03 	ext{ cm}^3$), the radius ($r = 10 	ext{ cm}$), and the problem explicitly tells us to use the approximation $\pi = 3$. Let's substitute these values to find the height ($h$), which represents the thickness of the monolayer:
$$0.03 = 3 	imes (10)^2 	imes h$$
$$0.03 = 300 	imes h$$
$$h = \frac{0.03}{300} = 1 	imes 10^{-4} 	ext{ cm}$$

### The Final Conversion Trap

We have successfully calculated the height as $10^{-4} 	ext{ cm}$. However, if you rush to the options, you might be tempted to pick option (b). **Stop and look at the units!**

All the options are given in meters ($	ext{m}$). This is a classic trap. We must convert our answer from centimeters to meters. Since $1 	ext{ cm} = 10^{-2} 	ext{ m}$, we multiply our result:
$$h = 10^{-4} 	imes 10^{-2} 	ext{ m} = 10^{-6} 	ext{ m}$$

Thus, the correct height of the monolayer is $10^{-6} 	ext{ m}$, making option (a) the correct choice. Always double-check your final units before marking the answer!

Select Answer:

Visualized Solution

\text{Understanding the Setup}

\text{Mass of Fatty Acid in } 10 \text{ mL}

\text{Calculating Mass}

\text{Volume of the Monolayer}

\text{Calculating Volume}

\text{Geometry of the Monolayer}

\text{Setting up the Equation}

\text{Calculating Height in cm}

\text{Converting to Meters}

\text{Conclusion \& Extensions}

The Sigma Insight: Molecular Mass, Mole Concept and Concentration

The Setup

The Mass and Volume Connection

The Geometry of the Monolayer

The Final Conversion Trap

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