Q42 of 44 Page 9

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?


(a) Oleic acid is hydrophobic (water-hating) in nature and thus it cannot be dissolved in water. Therefore, to reduce its concentration it is dissolved in alcohol in which it is highly soluble. This alcohol is then dissolved in water to form a thin layer of oleic acid, and since it is hydrophobic the oleic acid molecules get organized next to each other in a single layer which is one molecule thick.

(b) Lycopodium powder is used to distinguish the region of oleic acid and water by forming a boundary between the two. It also weakens the surface tension of the oleic acid and allows the layer to expand to its maximum size.


(c) Given:


Volume of oleic acid in given solution = 1 ml


Volume of alcohol in given solution = 19 ml


Volume of additional alcohol added in the solution = 20 ml


We know that, 1 ml of oleic acid is present in 20 ml of the solution. Therefore, in 1ml of solution,



Therefore, in every 1 ml of solution there would be 0.05 ml of oleic acid. Now since this 1 ml solution is diluted by adding further 20 ml of alcohol, the final solution,



(d) For determining the volume of n drops of this solution an instrument called burette will be used. The burette will be filled to a desired volume as marked on the burette scale. The burette allows the solution to be released drop by drops and thus n drops will be counted, the final value of the volume will be measure through burette scale and then subtracted from the initial value which will be the volume of the no of drops.


Volume of n drops = Initial value – Final value


(e) Let there be n drops in 1 ml of this solution which will be measured by using the burette. We know that in 1 ml of this solution there is 0.0025 ml of oleic acid, then,




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40

If velocity of light c, Planck’s constant h and gravitational constant G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

 41

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T =


where k is a dimensionless constant and g is acceleration due to gravity.


 43

(a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° fromthe earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.


(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.


(Comment : This is to illustrate why a telescope can magnify planets but not stars.)


 44

Einstein’s mass - energy relation emerging out of his famous theory of relativity relates mass (m ) to energy (E ) as E = mc2, where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where 1 MeV= 1.6×10-13J; the masses are measured in unified atomic mass unit (u) where

1u = 1.67 × 10-27 kg.


(a) Show that the energy equivalent of 1 u is 931.5 MeV.


(b) A student writes the relation as 1 u = 931.5 MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.