Surface Chemistry Test

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Surface Chemistry Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

Oil soluble dye is mixed with water in oil emulsion, then:

A
dispersion medium is coloured

B
dispersed phase is coloured

C
both are coloured

D
none is coloured

Solution

When an oil-soluble or water-soluble dye is added to the emulsion and observed under a microscope:

O/W Emulsion :
Water-soluble dye (amaranth): Continuous phase is colored and disperse phase is colorless.
Oil-soluble dye (scarlet red): Disperse phase is colored and the continuous phase is colorless.

W/O Emulsion :
Water-soluble dye (amaranth): Disperse phase is colored and the continuous phase is colorless.
Oil-soluble dye (scarlet red): Continuous phase is colored and disperse phase is colorless.
Question 2
1 / -0

One gram of charcoal adsorbs $$400$$ mL of $$0.5\ M$$ acetic acid to form a monolayer, and the molarity of acetic acid reduces to $$0.49\ M$$. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid, where the surface area of charcoal is $$3.01\times 10^{2}$$ m$$^{2}$$ g$$^{-1}$$.

A
$$1\times 10^{-19}m^{2}$$

B
$$1.2\times 10^{-19}m^{2}$$

C
$$2.5\times 10^{-21}m^{2}$$

D
None of these

Solution

$$400mL$$ of $$(0.5-0.49=0.01)\ M$$ acetic acid contains $$\displaystyle 0.4 \times 0.01 =0.2$$ moles of acetic acid or $$\displaystyle 4\times 10^{-3} \times 6.023 \times 10^{23} = 2.409 \times 10^{21}$$ molecules of acetic acid.

Total surface area of 1 g of charcoal is $$\displaystyle 3.01\times 10^{2} \ m^{2}$$.

Thus, the surface area occupied by 1 molecule of acetic acid is $$\displaystyle \frac {3.01\times 10^{2}\ m^{2}}{2.409 \times 10^{21}}=1.2\times 10^{-19}\ m^{2}$$.

Hence, the correct option is $$\text{B}$$
Question 3
1 / -0

Preparation of Lyophobic sols by chemical method involves :

A
double decomposition

B
oxidation and reduction

C
hydrolysis

D
all of these

Solution

Preparation of Lyophobic sols by chemical method involves
A) Double decomposition: By double decomposition: A sol of arsenic sulphide is obtained by passing hydrogen sulphide through a cold solution of arsenious oxide in water.
_{B) Oxidation and reduction:}
(a) By oxidation: A colloidal solution of sulphur can be obtained by bubbling oxygen (or any other oxidising agent like $$SO_2$$, etc.) through a solution of hydrogen sulphide in water.

(b) By reduction: A number of metals such as silver, gold and platinum, have been obtained in the colloidal state by treating the aqueous solution of their salts, with a suitable reducing agent such as formaldehyde, phenylhydrazine, hydrogen peroxide, stannous chloride etc.
_{C) By Hydrolysis:}
By hydrolysis: Many salt solutions are rapidly hydrolysed by boiling dilute solutions of their salts. For example, ferric hydroxide and aluminium hydroxide sols are obtained by boiling solutions of the corresponding chlorides.
Question 4
1 / -0

A soap $$(C_{17}H_{35} COONa)$$ solution becomes a colloidal sol at a concentration of $$1.2\times 10^{3} M$$. On the average, $$2.4\times 10^{13}$$ colloidal particles are present in $$1 mm^3$$. What is the average number of stearate ions (in multiples of $$10^6$$) in one colloidal particle (micelle)?

A
30

B
40

C
50

D
60

Solution

Concentration $$\displaystyle = 1.2 \times 10^3 M$$
1 $$\displaystyle mm^3 $$ will contain $$\displaystyle \dfrac {1.2 \times 10^3 mol/dm^3}{10^6 mm^3/dm^3} = 1.2 \times 10^{-3}$$ moles of soap.
1 mole of soap $$\displaystyle = 6.023 \times 10^{23}$$ molecules
$$\displaystyle 1.2 \times 10^{-3}$$ moles of soap $$\displaystyle =1.2 \times 10^{-3} \times 6.023 \times 10^{23} = 7.227 \times 10^{20}$$ stearate ions.
The average number of stearate ions in one colloidal particle (micelle) $$\displaystyle =\dfrac {7.227 \times 10^{20}}{2.4\times 10^{13}}=3 \times 10^{7} $$
Question 5
1 / -0

Which of the following are cation exchange resin and anion exchange resins?

A
$$NH_4OH$$ , $$RCHO$$

B
$$RNH_3OH$$ , $$RCOOH$$ or $$RSO_3H$$

C
$$RNH_3Cl$$ , $$RSO_4H$$

D
None of these

Solution

$$\begin{array}{l}\text{ Anion and cation resins are used in ion exchange }\\\text{ process. Positive charge is anion resins and }\\\text{ negative charge is cation resins. }\\\text{ Cation exchange resin is cross linked polymer }\\\text{ with negative ion and anion exchange resin is }\\\text{ vice-versa }\end{array}$$
Question 6
1 / -0

Match the List I with List II and choose the correct option from the codes given below:
List - I List - II
(P) Peptisation (1) Sky in blue
(Q) Delta Formation (2) Gold number
(R) Tyndall effect (3) Coagulation
(S) Protection power of colloids (4) Adsorption of common ion

A
P-3, Q-4, R-1, S-2

B
P-2, Q-3, R-1, S-4

C
P-3, Q-1, R-4, S-2

D
P-4, Q-3, R-1, S-2

Solution

Peptisation: Phenomenon in which a precipitate is passed into colloidal form by adsorption of common ions on its surface.

Delta formation: When river water (sol) meets seawater (electrolyte) coagulation occurs and delta is formed.

Tyndall effect: Scattering of light rays by suspended particles in a colloid is called the Tyndall effect and the sky looks blue due to Tyndall effect.

Protection of colloids: Protective power of colloids is inversely related to the Gold number.

Hence, option $$(D)$$ is correct.
Question 7
1 / -0

A student takes about 4 ml of distilled water in four test tubes marked $$P, Q, R$$ and $$S$$. He then dissolves in each test tube, an equal amount of various salts, viz. sodium sulphate in $$P$$, potassium sulphate in $$Q$$, calcium sulphate in $$R$$ and magnesium sulphate in $$S$$. After that, he adds an equal amount of soap solution in each test tube. On shaking each of these test tubes well, he observes a good amount of lather (foam) in the test tube marked:

A
$$P$$ and $$Q$$

B
$$Q$$ and $$R$$

C
$$P, Q$$ and $$S$$

D
$$P, R$$ and $$S$$

Solution

Soaps are sodium or potassium salts of long chain fatty acids. With $$Ca$$ and $$Mg$$, soap forms insoluble precipitate called scum. However, sodium (P) and potassium (Q) sulphates help in the formation of lather.

Hence, option A is correct.
Question 8
1 / -0

Pick out the statement which is not relevant in the discussion of colloids:

A
sodium aluminium silicate is used in the softening of hard water

B
potash alum is used in shaving rounds and as a styptic in medicine

C
artificial rain is caused by throwing electrified sand on the clouds from an aeroplane

D
deltas are formed at a place when the river pours its water into the sea

Solution

The statement which is not relevant in the discussion of colloids is sodium aluminium silicate is used in the softening of hard water. Sodium aluminium silicate and the softening of hard water have nothing to do with colloids. It is actually the simple chemical substitution of calcium salts with zeolite so that calcium zeolite precipitates out, and hardness of water removes.
Question 9
1 / -0

Foam is formed when_______

A
A solid dispersed in a liquid

B
A liquid dispersed in a liquid

C
A gas dispersed in a liquid

D
A liquid dispersed in a gas

Solution
Question 10
1 / -0

A physical adsorption at low temperature may pass into chemisorption as:

A
Temprarure decrease

B
Temperature increase

C
After some time at same temperature

D
physical adsorption never change into chemisorption

Solution

As temperature increases bonds tend to break thus resulting in chemisorption.

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Re-Attempt Weekly Quiz Competition

	List - I		List - II
(P)	Peptisation	(1)	Sky in blue
(Q)	Delta Formation	(2)	Gold number
(R)	Tyndall effect	(3)	Coagulation
(S)	Protection power of colloids	(4)	Adsorption of common ion

Surface Chemistry Test - 53

Result

Surface Chemistry Test - 53

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SHARING IS CARING

Question 1

dispersion medium is coloured

dispersed phase is coloured

both are coloured

none is coloured

Solution

Question 2

$$1\times 10^{-19}m^{2}$$

$$1.2\times 10^{-19}m^{2}$$

$$2.5\times 10^{-21}m^{2}$$

None of these

Solution

Question 3

double decomposition

oxidation and reduction

hydrolysis

all of these

Solution

Question 4

30

40

50

60

Solution

Question 5

$$NH_4OH$$ , $$RCHO$$

$$RNH_3OH$$ , $$RCOOH$$ or $$RSO_3H$$

$$RNH_3Cl$$ , $$RSO_4H$$

None of these

Solution

Question 6

P-3, Q-4, R-1, S-2

P-2, Q-3, R-1, S-4

P-3, Q-1, R-4, S-2

P-4, Q-3, R-1, S-2

Solution

Question 7

$$P$$ and $$Q$$

$$Q$$ and $$R$$

$$P, Q$$ and $$S$$

$$P, R$$ and $$S$$

Solution

Question 8

sodium aluminium silicate is used in the softening of hard water

potash alum is used in shaving rounds and as a styptic in medicine

artificial rain is caused by throwing electrified sand on the clouds from an aeroplane

deltas are formed at a place when the river pours its water into the sea

Solution

Question 9

A solid dispersed in a liquid

A liquid dispersed in a liquid

A gas dispersed in a liquid

A liquid dispersed in a gas

Solution

Question 10

Temprarure decrease

Temperature increase

After some time at same temperature

physical adsorption never change into chemisorption

Solution

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