10 \:\:mL of  1\:\: mM surfactant solution forms a monolayer covering 0.24\:\:cm^{2} on a polar substrate . If the polar head is approximated as a cube , what is its edge length ?

  • Option 1)

    1.0\:pm

  • Option 2)

    2.0\:pm

  • Option 3)

    0.1\:nm

  • Option 4)

    2.0\:nm

Answers (1)

 

Mole Concept -

One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.

- wherein

one mole = 6.0221367 X 1023   

 

 

 

Number of Moles -

No of moles = given mass of substance/ molar mass of substance

-

 

 

 

 

 

 

given 1\:\:mM\:\:and\:\:10\:\:mL

so

No.\:\: of\:\: moles\:\: absorbed = 10^{-3}\times10^{-3}\times10=10^{-5}

No.\:\: of\:\: molecules\:\: absorbed = 10^{-5}\times N_{A}

and    

total.\:\: area \:occupied\: = 10^{-5}\times N_{A}\times a^{2}

given 

10^{-5}\times N_{A}\times a^{2}=0.24(adsorbed \:\:area)

a^{2}=\frac{0.24}{10^{-5}\times N_{A}}cm^{2}

a^{2}=\frac{0.24}{10^{-5}\times 6\times 10^{23}}=4\times10^{20}cm^{2}

a=2\times10^{-12}m=2\:\:pm

  


Option 1)

1.0\:pm

Option 2)

2.0\:pm

Option 3)

0.1\:nm

Option 4)

2.0\:nm

Preparation Products

Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
Buy Now
Test Series JEE Main July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 17999/- ₹ 11999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 19999/-
Buy Now
Exams
Articles
Questions