Some Basic Concepts of Chemistry 2 Question 3
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3. $0.27 \mathrm{~g}$ of a long chain fatty acid was dissolved in $100 \mathrm{~cm}^{3}$ of hexane. $10 \mathrm{~mL}$ 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 $10 \mathrm{~cm}$. What is the height of the monolayer? [Density of fatty acid $=0.9 \mathrm{~g} \mathrm{~cm}^{-3} ; \pi=3$ ]
======= ####3. $0.27 \mathrm{~g}$ of a long chain fatty acid was dissolved in $100 \mathrm{~cm}^{3}$ of hexane. $10 \mathrm{~mL}$ 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 $10 \mathrm{~cm}$. What is the height of the monolayer? [Density of fatty acid $=0.9 \mathrm{~g} \mathrm{~cm}^{-3} ; \pi=3$ ]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $10^{-6} \mathrm{~m}$
(b) $10^{-4} \mathrm{~m}$
(c) $10^{-8} \mathrm{~m}$
(d) $10^{-2} \mathrm{~m}$
(2019 Main, 8 April II)
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Solution:
- $100 \mathrm{~mL}\left(\mathrm{~cm}^{3}\right)$ of hexane contains $0.27 \mathrm{~g}$ of fatty acid. In $10 \mathrm{~mL}$ solution, mass of the fatty acid,
$$ m=\frac{0.27}{100} \times 10=0.027 \mathrm{~g} $$
Density of fatty acid, $d=0.9 \mathrm{~g} \mathrm{~cm}^{-3}$
$\therefore$ Volume of the fatty acid over the watch glass,
$$ V=\frac{m}{d}=\frac{0.027}{0.9}=0.03 \mathrm{~cm}^{3} $$
Let, height of the cylindrical monolayer $=h \mathrm{~cm}$
$\because$ Volume of the cylinder $=$ Volume of fatty acid
$$ \begin{aligned} & \ \Rightarrow \quad & \ \Rightarrow \quad & =\pi r^{2} \times h \ \ldots & =\frac{V}{\pi r^{2}}=\frac{0.03 \mathrm{~cm}^{3}}{3 \times(10)^{2} \mathrm{~cm}^{2}} \ & =1 \times 10^{-4} \mathrm{~cm} \quad=1 \times 10^{-6} \mathrm{~m} \end{aligned} $$
