In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in 19 mL of alcohol. Then 1 mL of this solution is diluted to 20 mL by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule. Read the passage carefully and answer the following questions:

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of lycopodium powder?

(c) What would be the volume of oleic acid in each mL of solution prepared?

(d) How will you calculate the volume of n drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a long answer type question as classified in NCERT Exemplar

    (a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.

    (b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls.

    (c) In each mL of solution volume of oleic acid = 1/20mL × 1 20 = 1/400mL

    (d) Volume of n drops of this solution of oleic acid can be calculated by burette and measuring cylinder

    (e) If n drops of

    ...more

Similar Questions for you

V
Vishal Baghel

According to question, we can write

Total moles of gas = n = nOxygen + nOxygen = 1 6 2 + 1 2 8 3 2 = 1 2 m o l e s  

Volume of gas = 12 × 22.4 litre = 268.8 litre = 2.688 × 105 cm3 2 7 × 1 0 4 c m 3  

V
Vishal Baghel

K = Q r Δ x Δ A T

M L 2 T - 2 ( L ) L 2 ( θ ) ( T ) M 1 L 1 - T - 3 θ - 1

V
Vishal Baghel

Viscosity = Pascal . Second

P X A Y T Z = [ M 1 L 1 T 1 ]        

[ M 1 L + 1 T 1 ] x [ L 2 ] y [ T 1 ] z = M 1 L 1 T 1

= x = 1,     x + 2y = -1,      -x + z = 1

y = -1,              Z = 0

Viscosity = [P1A-1T0]

P
Payal Gupta

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10-2 cm

V
Vishal Baghel

Δ Q = m s Δ T s = Δ Q m Δ T [ s ] = M L 2 T 2 M Q = L 2 T 2 Q 1

Δ Q = m L [ L ] = M L 2 T 2 M = L 2 T 2

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 65k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1800k Answers

Learn more about...

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.

Need guidance on career and education? Ask our experts

Characters 0/140

The Answer must contain atleast 20 characters.

Add more details

Characters 0/300

The Answer must contain atleast 20 characters.

Keep it short & simple. Type complete word. Avoid abusive language. Next

Your Question

Edit

Add relevant tags to get quick responses. Cancel Post