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Step-by-Step Solution
Step 1: Understand the Problem
We have a dilute solution of oleic acid with a concentration of 0.01 cm3 of oleic acid per cm3 of solution. We create a thin film (monomolecular thickness) of area 4 cm2 by spreading 100 spherical drops of the solution. We are given the radius of each drop as
\left( \frac{3}{40\pi} \right)^{\frac{1}{3}} \times 10^{-3} \text{ cm} .
We need to find the thickness of the pure oleic acid layer in the form of
x \times 10^{-14}\text{ m} , and determine the value of x .
Step 2: Calculate the Total Volume of Oleic Acid (Undiluted) in the Drops
First, calculate the total volume of oleic acid in the 100 drops, considering the volume of each drop and the volume fraction. The volume of one spherical drop is:
V_{\text{drop}} = \frac{4}{3} \pi r^{3} ,
where r = \left( \frac{3}{40\pi} \right)^{\frac{1}{3}} \times 10^{-3} \text{ cm} .
The total volume of the 100 drops is then:
V_{\text{total}} = 100 \times \frac{4}{3} \pi r^{3}.
Step 3: Substitute the Given Radius and Simplify
Substitute
r = \left( \frac{3}{40\pi} \right)^{\frac{1}{3}} \times 10^{-3} \text{ cm}
into the expression for the total volume:
V_{\text{total}}
= 100 \times \frac{4}{3} \pi
\left( \left(\frac{3}{40\pi}\right)^{\frac{1}{3}} \times 10^{-3} \right)^{3}.
Since
\left(\frac{3}{40\pi}\right)^{\frac{1}{3}}
cubed is
\frac{3}{40\pi} ,
and
(10^{-3})^3 = 10^{-9} ,
we get:
V_{\text{total}} = 100 \times \frac{4}{3}\pi \times \frac{3}{40\pi} \times 10^{-9}.
Simplify this step by step:
The factor \frac{3}{40\pi} cancels partially with \frac{4}{3}\pi to yield \frac{4}{3}\pi \times \frac{3}{40\pi} = \frac{4}{40} = \frac{1}{10} .
Multiplying by 100 and by 10^{-9} gives:
100 \times \frac{1}{10} \times 10^{-9} = 10^{-8} \text{ cm}^3.
Step 4: Relate the Volume to the Thickness of the Film
This total volume 10^{-8} \text{ cm}^3 of oleic acid, when spread over an area of
4 cm2, will have a certain thickness (let’s call it t_{T} ). We write:
\text{Area} \times t_{T} = \text{Volume of oleic acid layer (pure)}.
Hence,
4 \times t_{T} = 10^{-8} \text{ cm}^3
\ \Rightarrow\ t_{T} = \frac{10^{-8}}{4} \text{ cm}^3/\text{cm}^2
= 2.5 \times 10^{-9} \text{ cm}.
However, carefully following the solution's derivation, the final step shows
t_{T} = 25 \times 10^{-10} \text{ cm},
which indeed is
2.5 \times 10^{-9} \text{ cm},
but often factored differently.
Convert t_{T} to meters:
t_{T} = 25 \times 10^{-10} \text{ cm}
= 25 \times 10^{-12} \text{ m}
= 2.5 \times 10^{-11} \text{ m}.
Step 5: Account for the Dilution Factor
The solution was 0.01 cm3 of oleic acid per cm3 of the solution. This means the actual thickness of the pure oleic acid layer t_{0} will be 0.01 of t_{T} :
t_{0} = 0.01 \times t_{T}
= 0.01 \times 25 \times 10^{-12} \text{ m}
= 25 \times 10^{-14} \text{ m}.
Step 6: Identify the Value of x
The thickness is expressed as x \times 10^{-14} \text{ m} . Comparing this to
25 \times 10^{-14} \text{ m},
we see that:
x = 25.
Answer
The value of x is 25.
