Some Basic Concepts of Chemistry - Result Question 3
3. 0.27 g of a long chain fatty acid was dissolved in $100 \mathrm{~cm}^3$ 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 $=0.9 \mathrm{~g} \mathrm{~cm}^{-3} ; \pi=3$ ]
(2019 Main, 8 April II)
(a) $10^{-6} \mathrm{~m}$
(b) $10^{-4} \mathrm{~m}$
(c) $10^{-8} \mathrm{~m}$
(d) $10^{-2} \mathrm{~m}$
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Answer:
Correct Answer: 3. (a)
Solution: $100 \mathrm{~mL}\left(\mathrm{~cm}^3\right)$ of hexane contains 0.27 g of fatty acid. In 10 mL solution, mass of the fatty acid,
$ m=\dfrac{0.27}{100} \times 10=0.027 \mathrm{~g} $
Density of fatty acid, $d=0.9 \mathrm{~g} \mathrm{~cm}^{-3}$
$\therefore$ Volume of the fatty acid over the watch glass,
$ V=\dfrac{m}{d}=\dfrac{0.027}{0.9}=0.03 \mathrm{~cm}^3 $
Let, height of the cylindrical monolayer $=h \mathrm{~cm}$
$\because$ Volume of the cylinder $=$ Volume of fatty acid
$\begin{aligned} \Rightarrow & V=\pi r^2 \times h \\ \Rightarrow & h=\dfrac{V}{\pi r^2}=\dfrac{0.03 \mathrm{~cm}^3}{3 \times(10)^2 \mathrm{~cm}^2} \\ & =1 \times 10^{-4} \mathrm{~cm}=1 \times 10^{-6} \mathrm{~m}\end{aligned}$
BETA
