Surface Chemistry Questions — JEE Chemistry

Q1

The formation of gas at the surface of tungsten due to adsorption is the reaction of order :

A 0

B 1

C 2

D insufficient data

Correct Answer

Option A

Solution

It is zero order reaction [NOTE : Adsorption of gas on metal surface is of zero order]

Q2

An example of solid sol is :

A Butter

B Hair cream

C Gem stones

D Paint

Correct Answer

Option C

Solution

An example of solid sol is gem stones.

Q3

A colloidal system consisting of a gas dispersed in a solid is called a/an :

A foam

B aerosol

C solid sol

D gel

Correct Answer

Option C

Solution

A colloidal system consisting of a gas dispersed in a solid is called a ‘solid sol’.

Q4

Which of the following is a correct statement?

A Brownian motion destabilises sols.

B Any amount of dispersed phase can added to emulsion without destablising it.

C Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids.

D Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution.

Correct Answer

Option D

Solution

As equal & similar charge particle will repel each other, hence will never precipitate.

Q5

Which of the following salt solutions would coagulate the colloid solution formed when

\mathrm{FeCl_3}

is added to

\mathrm{NaOH}

solution, at the fastest rate?

A 10 mL of 0.15 mol dm

^{-3}

CaCl

_2

B 10 mL of 0.2 mol dm

^{-3}

AlCl

_3

C 10 mL of 0.1 mol dm

^{-3}

Na

_2

SO

_4

D 10 mL of 0.1 mol dm

^{-3}

Ca

_3

(PO

_4

)

_2

Correct Answer

Option B

Solution

In the coagulation of a negative sol, the flocculating power is in the order. $\therefore \mathrm{Al}^{3+}>\mathrm{Ba}^{2+}>\mathrm{Na}^{+}$ and $\mathrm{FeCl}_{3}$ with $\mathrm{NaOH}$ forms a negative sol. $\therefore \mathrm{AlCl}_{3}$ coagulate it most.

Q6

If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log

{x \over m}

versus log P is linear. The slope of the plot is : (n and k are constants and n > 1)

A 2 k

B log k

C n

D

{1 \over n}

Correct Answer

Option D

Solution

According to Freundlich adsorption isotherm, in the median range of pressure

{x \over m} \propto {P^{{1 \over n}}}

$\Rightarrow$

{x \over m}

= kP

^{{1 \over n}}

taking log both sides, we get, log

{x \over m}

= logk +

{1 \over n}

logP Here in graph between log

{x \over m}

and logP, slope is

{1 \over n}

and intercepts = log k.

Q7

'Adsorption' principle is used for which of the following purification method?

A Chromatography

B Sublimation

C Distillation

D Extraction

Correct Answer

Option A

Solution

Principle used in chromotography is adsorption.

Q8

Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively. The correct order of their protective powers is

A D < A < C < B

B C < B < D < A

C A < C < B < D

D B < D < A < C

Correct Answer

Option C

Solution

For a protective colloid $\mu$ lesser the value of gold number better is the protective power.

Thus the correct order of protective power of

A,B,C

and

D

is $\Rightarrow$

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

\left( A \right)\,\, < \,\,\left( C \right)\,\, < \,\,\left( B \right)\,\, < \,\,\left( D \right)

Gold number

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.50\,\,\,\,\,\,0.10\,\,\,\,\,\,0.01\,\,\,\,\,\,0.005

Hence

(c)

is the correct answer

Q9

Haemoglobin and gold sol are examples of :

A positively charged sols

B negatively charged sols

C Positively and negatively charged sols, respectively

D negatively and positively charged sols, respectively

Correct Answer

Option C

Solution

Haemoglobin

\overset{\,}\longrightarrow

positive sol Ag $-$ sol

\overset{\,}\longrightarrow

negative sol

Q10

Which of the following is not an example of heterogenous catalytic reaction?

A Ostwald's process

B Combustion of coal

C Hydrogenation of vegetable oils

D Haber's process

Correct Answer

Option B

Solution

Then is no catalyst is required for combustion of coal.