Surface Chemistry — Page 4 | JEE Chemistry

Q31

Four gases A, B, C and D have critical temperatures

5.3,33.2,126.0

and

154.3 \mathrm{~K}

respectively. For their adsorption on a fixed amount of charcoal, the correct order is:

A D > C > B > A

B C > B > D > A

C D > C > A > B

D C > D > B > A

Correct Answer

Option A

Solution

The extent of adsorption of a gas on a solid surface depends on the critical temperature of the gas and the temperature of the solid surface.

The critical temperature is the temperature above which a gas cannot be liquefied, no matter how much pressure is applied.

Therefore, gases with higher critical temperatures are less likely to be adsorbed on a solid surface than gases with lower critical temperatures.

In this case, the gases are A, B, C, and D, with critical temperatures of 5.3 K, 33.2 K, 126.0 K, and 154.3 K, respectively.

Therefore, the correct order of adsorption on a fixed amount of charcoal is: D > C > B > A

Q32

Ferric chloride is applied to stop bleeding because -

A

\mathrm{FeCl}_{3}

reacts with the constituents of blood which is a positively charged sol.

B Blood absorbs

\mathrm{FeCl}_{3}

and forms a complex.

C

\mathrm{Cl}^{-}

ions cause coagulation of blood.

D

\mathrm{Fe}^{3+}

ions coagulate blood which is a negatively charged sol.

Correct Answer

Option D

Solution

The correct answer is Option D: Fe3+ ions coagulate blood which is a negatively charged sol.

Blood is a colloidal solution, which means that it is made up of small particles suspended in a liquid.

The particles in blood are negatively charged, and the negative charge prevents the particles from clumping together.

When ferric chloride is applied to a bleeding wound, the Fe3+ ions in the ferric chloride react with the negatively charged particles in the blood.

This reaction causes the particles to clump together, forming a clot that stops the bleeding.

The other options are incorrect: Option A is incorrect.

Blood is not a positively charged sol.

Option B is incorrect.

Blood does not absorb ferric chloride.

Option C is incorrect.

Cl- ions do not cause coagulation of blood.

Q33

Match the catalysts (Column I) with products (Column II). .tg .tg Column I Column II

List - I		List - II
(A)	V2O5	(i)	Polyethylene
(B)	TiCl4/Al(Me)3	(ii)	ethanal
(C)	PdCl2	(iii)	H2SO4
(D)	Iron Oxide	(iv)	NH3

A (A)-(iii); (B)-(i); (C)-(ii); (D)-(iv)

B (A)-(ii); (B)-(iii); (C)-(i); (D)-(iv)

C (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)

D (A)-(iv); (B)-(iii); (C)-(ii); (D)-(i)

Correct Answer

Option A

Solution

(A) V2O5 is used for preparation of H2SO4 in contacts process.

(B) TiCl4 + Al(Me)3 this mixture is called Ziegler-Natta catalyst which is used in the synthesis of high density polyethene from ethene.

(C) PdCl2 is used in ethanol formation (D) Iron oxide is used in Haber process for NH3.

Q34

Given .tg .tg Gas H2 CH4 CO2 SO2 Critical Temperature / K 33 190 304 630 On the basis of data given above, predict which of the following gases shows least adsorption on a definite amount of charcoal ?

A SO2

B CH4

C H2

D CO2

Correct Answer

Option C

Solution

Adsorption $\propto$ Tc Therefore, H2 gas shows least adsorption on a definite amount of charcoal.

Q35

Match with . (g) + 5{O_2}(g) \to 4NO(g) + 6{H_2}O(g)$$

List - I		List - II
(B)	${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$	(I)	$NO(g)$
(C)	${C_{12}}{H_{22}}{O_{11}}(aq) + {H_2}O(l) \to \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose}$	(II)	${H_2}S{O_4}(l)$
(D)	$2S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g)$	(III)	$Pt(s)$
()		(IV)	$Fe(s)$

A

(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})

B

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})

C

(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})

D (A) - (III), (B) - (II), (C) - (IV), (D) - (I)

Correct Answer

Option C

Solution

(A) $4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}(\mathrm{s})}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ (B) $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \stackrel{\mathrm{Fe}(\mathrm{s})}{\longrightarrow} 2 \mathrm{NH}_{3}(\mathrm{~g})$ (C) $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{l})}{\longrightarrow}$ $\underset{\text{ Glucose }}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}+\underset{\text{ Fructose }}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}$ (D) $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{NO}(\mathrm{g})}{\longrightarrow} 2 \mathrm{SO}_{3}$

Q36

Given below are the critical temperatures of some of the gases: .tg .tg Gas Critical temperature (K) He 5.2 CH

_4

190.0 CO

_2

304.2 NH

_3

405.5 The gas showing least adsorption on a definite amount of charcoal is

A He

B CH

_4

C CO

_2

D NH

_3

Correct Answer

Option A

Solution

The greater the value of critical temperature greater is adsorption as He has the least critical temperature so it adsorb least.

Q37

Match List I with List II .tg .tg LIST I LIST II A. Physisorption I. Single Layer Adsorption B. Chemisorption II.

20-40 \mathrm{~kJ} \mathrm{~mol}^{-1}

C.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \stackrel{\mathrm{Fe}(\mathrm{s})}{\longrightarrow} 2 \mathrm{NH}_{3}(\mathrm{~g})

III. Chromatography D. Analytical Application or Adsorption IV. Heterogeneous catalysis Choose the correct answer from the options given below:

A A - III, B - IV, C - I, D - II

B A - II, B - III, C - I, D - IV

C A - II, B - I, C - IV, D - III

D A - IV, B - II, C - III, D - I

Correct Answer

Option C

Solution

(A) Physisorption = 20 – 40 kJ/mol and Chemisorption = 80 – 240 kJ/mol (B) Physisorption is multi-layered and chemisorption is unimolecular layered.

(C) In heterogeneous catalysis, medium and catalyst are in different phases.

(D) Chromatography uses adsorption to purify/separate mixtures.

Q38

Match the following : .tg .tg

List - I		List - II
(a)	smoke	(i)	Foam
(b)	cell fluid	(ii)	Gel
(c)	jellies	(iii)	Aerosol
(d)	rubber (e) froth (f) Milk	(iv)	Emulsion

A (i)-(e), (ii)-(c), (iii)-(a), (iv)-(f)

B (i)-(d), (ii)-(b), (iii)-(a), (iv)-(e)

C (i)-(d), (ii)-(b), (iii)-(e), (iv)-(f)

D (i)-(b), (ii)-(c), (iii)-(e), (iv)-(d)

Correct Answer

Option A

Solution

Foam $\to$ Froth, whipped cream, soaplather Gel $\to$ Cheese, butter, jellies Aerosol $\to$ smoke dust Emulsion $\to$ milk

Q39

Match with : {O_3}\,.\,x{H_2}O$$

List - I		List - II
(B)	macromolecular colloid	(II)	CdS sol
(C)	positively charged sol	(III)	Starch
(D)	Cheese	(IV)	a gel

A (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

B (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

C (A) - (II), (B) - (III), (C) - (I), (D) - (IV)

D (A) - (I), (B) - (III), (C) - (II), (D) - (IV)

Correct Answer

Option C

Solution

Negative charged sol = CdS sol (II) Macromolecular colloid = starch (III) Positively charged sol = Fe2O3.xH2O (I) Cheese = gel (IV)

Q40

In Langmuir’s model of adsorption of a gas on a solid surface

A the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered

B the adsorption at a single site on the surface may involve multiple molecules at the same time

C the mass of gas striking a given area of surface is proportional to the pressure of the gas

D the mass of gas striking a given area of surface is independent of the pressure of the gas

Correct Answer

Option C

Solution

According to Langmuir's Model of adsorption of gas on a solid surface the mass of gas adsorbed(x) per gram of the adsorbent (m) is directly proportional to the pressure of the gas

(p)

at constant temperature i.e.

{x \over m} \propto p

$