10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm^2 on a polar substrate. If the

Q: Why do we assume the area covered by one molecule is $a^2$ and not the total surface area of the cube ($6a^2$)?

When a surfactant forms a monolayer, only one face of the cubic polar head is in contact with the polar substrate. The other faces are either touching neighboring molecules or facing the non-polar tail. Therefore, the footprint it leaves on the substrate is just $a^2$.

Q: Why did we use $6 \times 10^{23}$ for Avogadro's number instead of $6.022 \times 10^{23}$?

In many JEE problems, especially when the given values (like $0.24$) are multiples of 6, using $6 \times 10^{23}$ simplifies the calculation significantly without causing a major deviation in the final answer, which is often required to be matched with the given options.

Q: How do we convert the final answer from centimeters to picometers?

Recall that $1 \text{ cm} = 10^{-2} \text{ m}$ and $1 \text{ pm} = 10^{-12} \text{ m}$. Therefore, $1 \text{ cm} = 10^{10} \text{ pm}$. Multiplying $2 \times 10^{-10} \text{ cm}$ by $10^{10}$ gives exactly $2 \text{ pm}$.

Q: What does a '1 mM' solution mean?

'mM' stands for millimolar, which means $10^{-3}$ moles per liter (or $10^{-3}$ M). It's a measure of concentration.

Q: Could the polar head be a sphere instead of a cube?

In reality, molecules are not perfect cubes. However, for the sake of mathematical modeling and estimating dimensions, approximating them as cubes or spheres is a common technique in physical chemistry. The problem explicitly instructs us to assume a cubic shape.

Question

10 mL of 1 mM surfactant solution forms a monolayer covering $0.24 	ext{ cm}^2$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

Options:

(A) 2.0 pm

(B) 0.1 nm

(C) 1.0 pm

(D) 2.0 nm

Sigma Percentile · Accepted Answer

### Visualizing the Monolayer

Imagine a calm pool of water. When you add a surfactant, its molecules rush to the surface. They align themselves perfectly, forming a single layer known as a monolayer.

In our problem, these molecules are standing on a polar substrate. The problem gives us a fascinating detail: the polar head of each molecule can be approximated as a tiny cube.

Our mission is to find the edge length of this cubic head. To do this, we need to figure out the exact area a single molecule occupies on the substrate.

### Counting the Molecules

Before we can find the area of one molecule, we need to know how many molecules we have in total. We are given $10 	ext{ mL}$ of a $1 	ext{ mM}$ (millimolar) surfactant solution.

First, let's calculate the number of moles. We multiply the molarity by the volume in liters.

$$ 	ext{Moles} = 10^{-3} 	ext{ mol L}^{-1} 	imes 10 	imes 10^{-3} 	ext{ L} $$

This simple multiplication gives us $10^{-5} 	ext{ moles}$ of surfactant.

Now, to find the total number of molecules, we multiply the moles by Avogadro's number. To keep the math clean, we'll use $6 	imes 10^{23}$.

$$ 	ext{Number of molecules} = 10^{-5} 	imes 6 	imes 10^{23} = 6 	imes 10^{18} $$

We have exactly $6 	imes 10^{18}$ molecules forming our monolayer!

### The Footprint of a Single Molecule

These $6 	imes 10^{18}$ molecules collectively cover a total area of $0.24 	ext{ cm}^2$ on the substrate.

To find the area occupied by just one molecule, we divide the total area by the total number of molecules.

$$ 	ext{Area per molecule} = \frac{0.24 	ext{ cm}^2}{6 	imes 10^{18}} $$

Dividing $0.24$ by $6$ gives $0.04$. So, the area of one molecule's footprint is $0.04 	imes 10^{-18} 	ext{ cm}^2$, which we can neatly rewrite as $4 	imes 10^{-20} 	ext{ cm}^2$.

### Unlocking the Edge Length

Here is where the geometry kicks in. The problem states that the polar head is a cube.

When a cube sits on a flat surface, the area it covers is simply the area of its square face. If the edge length of the cube is $a$, then the area of this face is $a^2$.

We can now equate this geometric area to the physical area we just calculated:

$$ a^2 = 4 	imes 10^{-20} 	ext{ cm}^2 $$

Taking the square root of both sides, we find the edge length $a$:

$$ a = 2 	imes 10^{-10} 	ext{ cm} $$

### The Final Conversion

Our answer is in centimeters, but the options are in picometers and nanometers. Let's convert it to picometers.

We know that $1 	ext{ cm} = 10^{-2} 	ext{ m}$ and $1 	ext{ pm} = 10^{-12} 	ext{ m}$. Therefore, $1 	ext{ cm}$ is equal to $10^{10} 	ext{ pm}$.

$$ a = 2 	imes 10^{-10} 	imes 10^{10} 	ext{ pm} = 2.0 	ext{ pm} $$

The edge length of the cubic polar head is exactly **$2.0 	ext{ pm}$**. This matches option (a) perfectly!

Animated Solution for Chemistry - States of Matter: 10 mL of 1 mM surfactant solution forms a monolayer covering $0.24 cm^{2}$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

Select Answer:

Visualized Solution

$Visualizing the Setup$

$The Strategy$

$Calculating Moles$

$Moles Result$

$Calculating Molecules$

$Molecules Result$

$Area per Molecule$

$Area Result$

$Relating Area to Edge Length$

$Final Calculation$

The Sigma Insight: Solid State

Visualizing the Monolayer

Counting the Molecules

The Footprint of a Single Molecule

Unlocking the Edge Length

The Final Conversion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Strategy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Moles

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Molecules

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Molecules Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area per Molecule

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area Result

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating Area to Edge Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation