Surface Chemistry(11th And 12th > Chemistry ) Questions and answers for exam Preparation

Question 1.

Which of the following statements are true about peptization? A. A colloidal sol is converted into precipitate B. It is done in the presence of a small electrolyte C. Charges develop on the precipitate during this process D. The charges developed during this process further breaks up into smaller particles of the size of colloids.

A,B,D
A,B,C
B,C,D
A,B,C,D

Explanation:-

Answer: Option C. -> B,C,D
:
C

During peptization, precipitate is converted into a colloidal sol by shaking it with dispersion medium in the presence of a small amount of electrolyte.

Question 2.

Which of the following properties of colloids involve scattering of lights?

Colligative properties
Tyndall effect
Brownian movement
None of the above

Explanation:-

Answer: Option B. -> Tyndall effect
:
B

The Tyndall effect is due to the fact that colloidal particles scatter light in all directions in space.

Question 3.

The volume of colloidal particle $V_{c}$ as comapred to the volume of solute particle in a true solution $V_{s}$ could be

$\frac{V_{c}}{V_{s}} = 1$
$\frac{V_{c}}{V_{s}} = 10^{23}$
$\frac{V_{c}}{V_{s}} = 10^{- 3}$
$\frac{V_{c}}{V_{s}} = 10^{3}$

Explanation:-

Answer: Option D. ->

\frac{V_{c}}{V_{s}} = 10^{3}

:
D

Diameter of colloidal particle is about 10 times larger than solute particles in a true solution. Hence, volume is about 1000 times greater i.e. $\frac{V_{c}}{V_{s}} = 10^{3}$

Question 4.

Zeta potential ( or electrokinetic potential ) is the

potential required to bring about coagulation of a colloidal sol
potential required to give the particles a speed of 1 cm/ sec in the sol
potential difference between fixed charged layer and diffused layer having opposite charge
potential energy of colloidal particles

Explanation:-

Answer: Option C. -> potential difference between fixed charged layer and diffused layer having opposite charge
:
C

Zeta potential is the potential difference between fixed charge layer and diffused layer having opposite charge

Question 5.

$10^{- 4}$ g of gelatin is required to be added to 100 $c m^{3}$ of a stanadard gold solution to just prevent its precipitation by addition of 1 $c m^{3}$ of 10% NaCl solution to it. Hence, gold number of gelatin is?

10
1
0.1
0.01

Explanation:-

Answer: Option D. -> 0.01
:
D

$10^{- 4} g = 0.1 m g$
$100 c m^{3}$ of gold sol require 0.1 mg
$∴ 10 c m^{3}$ of gold sol will required 0.01 mg
Hence, gold number = 0.01.

Question 6.

Which of the following does not affect the Brownian movement:

Nature of colloid
Size of the particles
Viscosity of the solution
None of the above

Explanation:-

Answer: Option A. -> Nature of colloid
:
A

This motion is independent of the nature of the colloid but depends on the size of the particles and viscosity of the solution.

Question 7.

Arrange the following in the decreasing order of their coagulation power: $A l^{3 +}, N a^{+}, C a^{2 +}$

$A l^{3 +} = C a^{2 +} = N a^{+}$
$N a^{+} > C a^{2 +} > A l^{3 +}$
$N a^{+} > A l^{3 +} > C a^{2 +}$
$A l^{3 +} > C a^{2 +} > N a^{+}$

Explanation:-

Answer: Option D. ->

A l^{3 +} > C a^{2 +} > N a^{+}

:
D

Greater the valency of the flocculating ion added, greater is its power to cause precipitation (Hardy Schulze rule).

Question 8.

Ferric chloride is added to stop bleeding because

$F e^{3 +}$ ions coagulate positively charged blood solution
$F e^{3 +}$ ions coagulate negatively charged blood solution
$C l^{-}$ ions coagulate positively charged blood solution
$C l^{-}$ ions coagulate negatively charged blood solution

Explanation:-

Answer: Option B. ->

F e^{3 +}

ions coagulate negatively charged blood solution
:
B

Negatively charged blood solution is coagulated by $F e^{3 +}$ ions.

Question 9.

Milk can be preserved by adding a few drops of

Formic acid solution
Formaldehyde solution
Acetic acid solution
Acetaldehyde solution

Explanation:-

Answer: Option B. -> Formaldehyde solution
:
B

Formaldehyde is an emulsifier

Question 10.

Silver iodide is used in bringing about artificial rain because silver iodide

is easily decomposed in presence of sunlight
contains iodide ions which are large in size
has crystal structure similar to ice and leads to coagulation
being heavy, its spray brings the water droplets down

Explanation:-

Answer: Option C. -> has crystal structure similar to ice and leads to coagulation
:
C

$A g^{+} I^{-}$ ions brings about coagulation of water droplets easily.