JEE Main · 2019 · Shift-IIhardMOLE-192

0.27\,g of a long chain fatty acid was dissolved in 100\,cm3 of hexane. 10\,mL of this solution was added dropwise to…

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

0.27 g0.27\,\text{g} of a long chain fatty acid was dissolved in 100 cm3100\,\text{cm}^3 of hexane. 10 mL10\,\text{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 cm10\,\text{cm}. What is the height of the monolayer? [Density of fatty acid =0.9 g cm−3= 0.9\,\text{g cm}^{-3}; π=3\pi = 3]

Options
  1. a

    10−4 m10^{-4}\,\text{m}

    ✓
  2. b

    10−6 m10^{-6}\,\text{m}

  3. c

    10−8 m10^{-8}\,\text{m}

  4. d

    10−2 m10^{-2}\,\text{m}

Correct Answera

10−4 m10^{-4}\,\text{m}

Detailed Solution

🧠 Watch-Glass Monolayer Geometry Imagine the fatty acid molecule as a tiny pencil. When you drop it on water, it stands up vertically and touches its neighbours, forming a carpet exactly one "pencil-height" (hh) thick. The volume of the carpet is simply its Area (AA) times its height (hh).

🗺️ Strategic Roadmap Step 1: Find the Aliquot Load. Stock: 0.27 g0.27\,g in 100 mL100\,mL. Aliquot (10 mL10\,mL) mass =0.027 g= 0.027\,g.

Step 2: Bulk Volume. Density =0.9 g/cm3= 0.9\,g/cm^3. V=0.027/0.9=0.03 cm3V = 0.027 / 0.9 = 0.03\,cm^3

Step 3: The "Carpet" Area. Watch glass radius =10 cm= 10\,cm. Area =πr2=3×100=300 cm2= \pi r^2 = 3 \times 100 = 300\,cm^2.

Step 4: Find the Height. h=V/A=0.03/300=0.0001 cm=10−4 cmh = V / A = 0.03 / 300 = 0.0001\,cm = 10^{-4}\,cm Converting to meters (mm): 10−6 m10^{-6}\,m.

\boxed{\text{Answer: (a) 10^{-6} m}}

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