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NEET • CHEMISTRY • SURFACE CHEMISTRY

Surface chemistry practice · Hard

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1 Hard

1 of 22

Application +4 | -1

A cationic colloidal electrolyte forms micelle at

10^{-3} \mathrm{M}

concentration in water. Its

1 \mathrm{~mm}^3

solution contains

2 \times 10^{13}

micelle structures. What is the number of cations contained in one micelle? Take

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

View answer and explanation

Correct option: A

Option A (30) is correct because it involves calculating the number of micelles and then dividing the total number of cations by the number of micelles.

2 Hard

2 of 22

Understanding +4 | -1

Most suitable salt which can be used for efficient clotting of blood will be

View answer and explanation

Correct option: D

FeC{l_3}

Option D (

FeC{l_3}

) is correct because, according to the Hardy-Schulze rule, ions with higher charges are more effective in causing coagulation.

Fe^{3+}

has a higher charge than

Fe^{2+}

Mg^{2+}

, or

Na^+

. Blood is a negatively charged sol, so a cation is needed.

3 Hard

3 of 22

Application +4 | -1

The density of gold is

19\;\mathrm{g/c}{\mathrm{m}^3}

. If

{1.9 \times {{10}^{ - 4}}\;\mathrm{g}}

of gold is dispersed in 1 L of water to give a sol having spherical gold particles of radius

10\;\mathrm{ }nm

, then the number of gold particles per

m{m^3}

of the sol will be

View answer and explanation

Correct option: D

2.4 \times {10^6}

Option D (

2.4 \times {10^6}

) is correct because it is calculated by finding the volume of a single gold particle, then the total number of particles, and finally the number of particles per unit volume.

4 Hard

4 of 22

Application +4 | -1

10 mL of mM surfactant solution forms a monolayer covering

0.24\,\,c{m^2}

on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

View answer and explanation

Correct option: A

2.0 pm

Option D (2.0 pm) is correct because it's calculated by finding the area occupied by one surfactant molecule and then taking the cube root to find the edge length.

5 Hard

5 of 22

Application +4 | -1

The coagulation of

100 \mathrm{~mL}

of a colloidal sol of gold is completely prevented by addition of 250

\mathrm{mg}

of a substance to it before adding

10 \mathrm{~mL}

1 \% \mathrm{NaCl}

solution. What is the gold number of this substance?

View answer and explanation

Correct option: A

Option A (25) is correct because the gold number is defined as the mass in milligrams of a protective colloid that prevents the coagulation of 10 mL of a gold sol when 1 mL of a 10% NaCl solution is added.

6 Hard

6 of 22

Understanding +4 | -1

Select the correct statements among the following

a) Foam is a colloidal solution of gas in liquid

b) Aerosol is a colloidal solution of liquid in gas

c) Gel is a colloidal solution of solid in liquid

d) Emulsion is a colloidal solution of liquid in liquid

View answer and explanation

Correct option: A

a, b, d correct

Option A (a, b, d correct) is correct because foam is gas in liquid, aerosol is liquid in gas, and emulsion is liquid in liquid. Gel is liquid in solid.

7 Hard

7 of 22

Application +4 | -1

The coagulation of

400 \mathrm{~mL}

of a positive colloid took place when

1.46 \mathrm{~g}

\mathrm{HCl}

was added to it without changing the volume much. What is the flocculation value of

\mathrm{HCl}

for the colloid?

View answer and explanation

Correct option: A

100

Option A (100) is correct because the flocculation value is the number of millimoles of electrolyte required to coagulate 1 liter of colloidal solution. We calculate the moles of HCl, convert to millimoles, and then adjust for the volume.

8 Hard

8 of 22

Application +4 | -1

10 \%

sites of catalyst bed have adsorbed by

\mathrm{H}_2

. On heating

\mathrm{H}_2

gas is evolved from sites and collected at

0.03 \mathrm{~atm}

and

300 \mathrm{~K}

in a small vessel of

2.46 \mathrm{~cm}^3

.
No. of sites available is

5.4 \times 10^{16}

per

\mathrm{cm}^2

and surface area is

1000 \mathrm{~cm}^2

. Find out the no. of surface sites occupied per molecule of

\mathrm{H}_2

\left(\right.

Given

\mathrm{N}_{\mathrm{A}}=6 \times 10^{23}

)

View answer and explanation

Correct option: A

Option A (3) is correct because it represents the number of surface sites occupied per molecule of H2, calculated from the given data.

9 Hard

9 of 22

Application +4 | -1

A sample of

16 \mathrm{~g}

charcoal was brought into contact with

\mathrm{CH}_4

gas contained in a vessel of 1 litre at

27^{\circ} \mathrm{C}

. The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is

1.6 \mathrm{~g} / \mathrm{cm}^3

. What is the volume of the

\mathrm{CH}_4

gas adsorbed per gram of the adsorbent at 608 torr and

27^{\circ} \mathrm{C}

View answer and explanation

Correct option: A

16.25

Option A (16.25) is correct because it represents the volume of CH4 adsorbed per gram of charcoal, calculated using the ideal gas law and the pressure drop.

10 Hard

10 of 22

Application +4 | -1

3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is

View answer and explanation

Correct option: A

18 mg

Option A (18 mg) is correct because it represents the amount of acetic acid adsorbed per gram of charcoal, calculated from the change in normality.

11 Hard

11 of 22

Application +4 | -1

In an adsorption experiment, a graph between

\log (\mathrm{x} / \mathrm{m})

versus

\log \mathrm{P}

was found to be linear with a slope of

45^{\circ}

. The intercept on the

y

axis was found to be 0.301 . The amount of the gas adsorbed per gram of charcoal under a pressure of

3.0 \mathrm{~atm}

View answer and explanation

Correct option: A

Option A (6) is correct because the amount of gas adsorbed per gram of charcoal is calculated using the Freundlich isotherm equation.

12 Hard

12 of 22

Application +4 | -1

50\mathrm{ }\;mL\mathrm{ }\;of\mathrm{ }\;1\;\mathrm{ }M

oxalic acid is shaken with

0.5\;g

wood charcoal. The final concentration of the solution after adsorption is

0.5\;\mathrm{ }M

. What is the amount of oxalic acid absorbed per gram of carbon?

View answer and explanation

Correct option: C

6.30\;\mathrm{ }g

Option C (6.30 g) is correct because it represents the amount of oxalic acid absorbed per gram of carbon, calculated from the change in concentration and volume.

13 Hard

13 of 22

Understanding +4 | -1

The Gold number of four protective colloids A, B, C and D are 0.03, 0.003, 10 and 30 resp. protective power of these colloids will be in the order

View answer and explanation

Correct option: A

B > A > C > D

Option A (B > A > C > D) is correct because the gold number is inversely proportional to the protective power of a colloid. A lower gold number indicates higher protective power.

14 Hard

14 of 22

Understanding +4 | -1

Amongst the following statements regarding adsorption, those that are valid are:
(A)

\Delta \,H

becomes less negative as adsorption proceeds.
(B) On a given adsorbent, ammonia is adsorbed more than nitrogen gas.
(C) On adsorption, the residual force acting along the surface of the adsorbent increases.
(D) With increase in temperature, the equilibrium concentration of adsorbate increases.

View answer and explanation

Correct option: C

(A) and (B)

Option C is correct.

\Delta H

becomes less negative as adsorption proceeds because fewer active sites are available, and ammonia is adsorbed more than nitrogen gas due to stronger interactions.

15 Hard

15 of 22

Understanding +4 | -1

Among the following, the false statement is

View answer and explanation

Correct option: D

Latex is a colloidal solution of rubber particles which are positively charged

Option D (Latex is a colloidal solution of rubber particles which are positively charged) is the false statement because latex particles are negatively charged, not positively charged.

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16 Hard

16 of 22

Understanding +4 | -1

The example of anionic surfactants is

View answer and explanation

Correct option: B

{C_{15}}{H_{31}}COONa

Option B (

{C_{15}}{H_{31}}COONa

) is correct because it is a soap, which is an anionic surfactant. When dissolved in water, it forms

{C_{15}}{H_{31}}COO^-

and

Na^+

, with the anion being the surface-active part.

17 Hard

17 of 22

Application +4 | -1

A detergent

\left(\mathrm{C}_{12} \mathrm{H}_{25} \mathrm{SO}_4^{-} \mathrm{Na}^{+}\right)

solution becomes a colloidal sol at a concentration of

10^{-3} \mathrm{M}

. On an average

10^{13}

colloidal particles are present in

1 \mathrm{~mm}^3

. What is the average number of ions which are contained by one colloidal particle (micelle)?

View answer and explanation

Correct option: A

Option A (60) is correct because it involves calculating the number of colloidal particles per liter and then dividing the number of detergent molecules per liter by the number of colloidal particles per liter to find the average number of ions per micelle.

18 Hard

18 of 22

Application +4 | -1

The density of gold is

19 \mathrm{~g} / \mathrm{cm}^3, 1.9 \times 10^{-3} \mathrm{~g}

of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius

10.5 \mathrm{~nm}

.
The gold particles are occupying

50 \%

of the available space. The number of gold particles per

\mathrm{mm}^3

is about

10^n

. Find the value of

n

View answer and explanation

Correct option: A

Option A (7) is correct because it involves calculating the volume of a single gold particle, the total number of particles, and then the number of particles per unit volume, considering the space occupied.

19 Hard

19 of 22

Understanding +4 | -1

One desires to prepare a positively charged sol of silver iodide. This can be achieved by

View answer and explanation

Correct option: B

Adding a little

\mathrm{KI}

solution to

\mathrm{AgNO}_3

solution in slight excess

Option B (Adding a little

\mathrm{KI}

solution to

\mathrm{AgNO}_3

solution in slight excess) is correct because when

\mathrm{AgNO}_3

is in excess, it adsorbs

Ag^+

ions, leading to a positively charged sol.

20 Hard

20 of 22

Understanding +4 | -1

A negatively charged sol can be formed by peptizing a solutions of

View answer and explanation

Correct option: B

{AgI}

with

{KI}

Option B (

{AgI}

with

{KI}

) is correct because when

{KI}

is in excess,

I^-

ions are adsorbed onto the AgI precipitate, forming a negatively charged sol.

21 Hard

21 of 22

Understanding +4 | -1

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) Photochemical smog causes cracking of rubber.
Reason (R) Presence of ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate in photochemical smog makes it oxidising.
Choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: B

Both (A) and (R) are true and (R) is the correct explanation of (A).

Option B is correct because both the assertion and reason are true, and the reason correctly explains why photochemical smog causes cracking of rubber due to its oxidizing nature.

22 Hard

22 of 22

Application +4 | -1

One gram of activated carbon has a surface area of

1000 \mathrm{~m}^2

. Considering complete coverage as well as monomolecular adsorption, how much ammonia at

1 \mathrm{~atm}

and

273 \mathrm{~K}

would be adsorbed on the surface of

\frac{44}{7} \mathrm{~g}

carbon if radius of a ammonia molecules is

10^{-8} \mathrm{~cm}

View answer and explanation

Correct option: A

7.46

Option A (7.46) is correct because it is the calculated amount of ammonia adsorbed based on the given surface area and molecular radius.

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Key Takeaways

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