JEE PYQ: Surface Chemistry Question 24
Question 24 - 2019 (08 Apr Shift 2)
Type: MCQ
(0). (27 g of a long chain fatty acid was dissolved in 100 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 $= 0.9$ g cm$^{-3}$; $\pi = 3$]
(a) $10^{-6}$ m
(b) $10^{-8}$ m
(c) $10^{-2}$ m
(d) $10^{-4}$ m
Show Answer
Answer: 0
Solution
Given: 0.27 g is present in 100 cm$^3$ of hexane $\therefore$ 10 mL of aqueous solution contains only 0.027 g acid.
Volume of 0.027 g acid $= \frac{0.027}{0.9} = 0.03$ mL
$\pi r^2 h = \frac{0.027}{0.9}$ (given $r = 10$ cm, $\pi = 3$)
$\therefore h = 10^{-4}$ cm $= 10^{-6}$ m
BETA
