It is highly specific in nature

It is caused by chemical bond formation

MCQ Questions for Class 12 Engineering Chemistry Surface Chemistry Quiz 11

Consider the following reactions at 300 K: X

$\rightarrow$ Y (uncatalysed reaction) & X

$\rightarrow$ Y (catalysed reaction). The energy of activation is lowered by 83.14 kJmol

$^{-1}$ for the catalysed reaction.

The rate of catalysed reaction is:

0%
$3\times 10^{14}$ times that of uncatalysed reaction
0%
$15\times 10^{14}$ times that of uncatalysed reaction
0%
25 times that of uncatalysed reaction
0%
22 times that of uncatalysed reaction

Explanation

The Arrhenius equation for the uncatalyzed reaction is

$\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{\iota \mathrm{R}}/\mathrm{R}\mathrm{T}}.$

The Arrhenius equation for the catalyzed reaction is

$\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}=\mathrm{A}\mathrm{e}^{-\mathrm{E}_{ \mathrm{R}}/\mathrm{R}\mathrm{T}}.$

Let A' be the activation energy for the uncatalyzed reaction.

The activation energy for the catalyzed reaction will be

$(\mathrm{A'}-83.14\times 10^{3}).$

Divide the two equations.

$\displaystyle \frac{\mathrm{K}_{\mathrm{c}\mathrm{a}\mathrm{t}}}{\mathrm{K}_{\mathrm{m}\mathrm{c}\mathrm{a}\mathrm{t}}}=\frac{\mathrm{e}^{-(\mathrm{A'}-83.14\times 10^{3})/\mathrm{R}\mathrm{T}}}{\mathrm{e}^{-\mathrm{A'}/\mathrm{R}\mathrm{T}}}=\mathrm{e}^{83.14\times 10^{3}/\mathrm{R}\mathrm{T}}=\mathrm{e}^{33.33} \approx 3 \times 10^{14}$

A reaction is catalysed by

$H^{+}$ ion in the presence of

$HA$ , rate constant is

$2\times 10^{-3}min^{-1}$ and in presence of

$HB$ , rate constant is

$1\times 10^{-3}min^{-1}$ .

$HA$ and

$HB$ (both strong acids) have relative strength as:

0%
$0.5$
0%
$0.002$
0%
$0.001$
0%
$2$

Explanation

The relative strength is the ratio of the rate constants.

$Relative\ strength = \dfrac {Rate\ constant\ of\ HA} {Rate\ constant \ of \ HB} = \dfrac {2\times 10^{-3}min^{-1}} {1\times 10^{-3}min^{-1}}=2.$

Methylene blue is adsorbed on activated charcoal at

$25^{0}C$ from its aqueous solution. Which is the correct statement for this process?

0%
The adsorption requires activation at $25^{0}C$ .
0%
The adsorption is accompanied by a decrease in enthalpy.
0%
The adsorption increases with increase of temperature.
0%
The adsorption is irreversible.

Which of the following are the correct statements regarding the catalyzed and uncatalyzed reactions at constant temperature?

0%
Catalyzed reaction does not alter $\Delta H$ of the reaction
0%
At constant temperature, $\Delta G^0$ of the catalysed and uncatalyzed reaction is same
0%
Catalyzed reaction alters equilibrium constant even at constant temperature
0%
Catalysed reaction alters $r_f$ and $r_b$ values equally at constant temperature

Explanation

The characteristics of a catalyst are as follows:

i) A catalyst remains unchanged chemically at the end of reaction, however, its physical state may change.

ii) A catalys can never initiate a chemical reaction. It simply influences the rate of reaction.

iii) A small quantity of catalyst is sufficient to influence the rate of reaction e.g.

$1$ g atom of Platinum is sufficient to catalyse

$108$ litres of

$H_2O_2$ decomposition.

iv) A catalyst does not influence the equilibrium constant of reaction. It simply helps in attaining equilibrium earlier. It alters the rate of forward & backward reactions equally.

v) Catalyst’s activity is more or less specific. A catalyst for one reaction is not necessary to catalyse the another reaction.

A catalyst does not alter

$\Delta H$ of the reaction. At constant temperature,

$\Delta G^o$ of the catalysed and uncatalysed reaction remains the same. Also, catalysed reaction alter

$r_f$ and

$r_b$ values equally at constant temperature.

A reaction is catalysed by

$H^+$ ion. In presence of HA, rate constant is

$2\times 10^{-3} min^{-1}$ and in presence of HB rate constant is

$1\times 10^{-3} min^{-1}$ , HA and HB both being strong acids, we may conclude that:

0%
equilibrium constant is 2.
0%
HA is stronger acid than HB.
0%
relative acidic strength of HA and HB is 2.
0%
HA is weaker acid than HB and relative strength is 0.5.

Explanation

Rate

$=$ k [reactant] [HA or HB].

Rate is more with HA so it is stronger than HB.

Relative strength of HA and HB

$=\frac {2\times 10^{-3}}{1\times 10^{-3}}=2$ .

Therefore options B and C are correct.

Assertion: A negative catalyst retards the rate of chemical reaction.
Reason: Presence of negative catalyst boosted up the energy of activation.

0%
Assertion is correct but Reason is wrong
0%
Assertion is wrong but Reason is correct
0%
Both Assertion and explanation are correct and Reason is correct explanation of Assertion
0%
Both Assertion and explanation are correct but Reason is not correct explanation of Assertion

The solution of

$H_2O_2$ can be stored for a long time at room temperature. However, bubbles of oxygen form as soon as a drop of bromine is added. The role of bromine for the reaction

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$ is :

0%
to act as a positive catalyst
0%
to form an intermediate $(Br^-)$ on reacting with $H_2O_2$ following reduction of bromine and then to show oxidation of $Br^{-}$ to $Br_2$ by $H_2O_2$
0%
to act as heterogeneous catalyst
0%
to act as negative catalyst

Explanation

In the decomposition of hydrogen peroxide, two successive reactions of

$H_2O_2$ , with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decomposition, both of them must have significantly lower activation energies than the uncatalyzed decomposition, as shown schematically in Figure.

Bromine acts as negative catalyst in the reaction. Bromine forms an intermediate

$(Br^-)$ on reacting with

$H_2O_2$ following reduction of bromine and then to show oxidation of

$Br^{-}$ to

$Br_2$ by

$H_2O_2$ .

Which acts as a negative catalyst?

0%
Lead tetraethyl as antiknock compound.
0%
Glycerol in decomposition of $H_2O_2$ .
0%
Ethanol in oxidation of chloroform.
0%
None of the above.

Consider the following statements and arrange in the order of true/false as given in the codes.

$S_1:$ The rate of the reaction A

$\rightarrow$ B having the rate law

$- \displaystyle \frac{d[A]}{dt} = k [A] [B]$ when plotted against time will exhibit a maximum at some time.

$S_2:$ A catalyst in a chemical reaction increases the forward E

$_a$ and decreases the backward

$E_a$ .

$S_3:$ A catalyst in a chemical reaction decreases both forward and backward

$E_a$ .

$S_4:$ For a first-order reaction, the time required to reduce successively the concentration of reactant by a constant fraction is always same.

0%
TTTT
0%
FFFF
0%
FTFT
0%
TFTT

Explanation

The rate of the reaction A

$\rightarrow$ B having the rate law

$-\cfrac{d[A]}{dt}=k[A][B]$ when plotted against time will exhibit a maximum at some time.

This is because, rate = 0 at t = 0 and t = t

$_\inf$ . Hence statement 1 is true.

A catalyst in a chemical reaction decreases the forward Ea and decreases the backward Ea.

Note: It decreases both by an equal amount. Hence, statement 2 is false.

A catalyst in a chemical reaction either increases or decreases both forward and backward Ea by an equal amount. Hence statement 3 is true.

For a first-order reaction, the time required to reduce successively the concentration of reactant by a constant fraction is always the same. This is because the half-life period is constant for the first-order reactions. Hence, statement (4) is correct.

Select the incorrect statement regarding the given reaction.

$RCOCl+H_2\xrightarrow{Pd-BaSO_4} RCHO+HCl$

0%
$Pd$ acts as a positive catalyst.
0%
It is Rosenmund reaction.
0%
$BaSO_4$ acts as poison for $Pd$ .
0%
$BaSO_4$ acts as promoter for $Pd$ .

Explanation

The Rosenmund reaction is catalyzed by palladium on barium sulphate. Barium sulphate reduces the activity of palladium due to its low surface area meaning it decreases the reducing power of palladium in order to prevent over-reduction of the acid.

So,

$BaSO_4$ acts as a poison for Pd, not as a promoter for Pd

So, D is here correct answer

Identify the true statement(s):

0%
A catalyst is chemically unchanged at the end of a reaction
0%
A catalyst may appear in the kinetic rate equation of the reaction
0%
A catalyst will not affect the composition of an equilibrium mixture
0%
A catalyst cannot cause a non-spontaneous ( $\Delta$ G > 0) reaction to proceed

The catalyst provides a new pathway involving lower amount of activation energy, to make the reaction fast. Which of the following is/are the result of catalytic action?

0%
Catalyst decreases $E_a/RT$ value.
0%
Catalyst increases $-E_a/RT$ value.
0%
Catalyst increases $e^{E_a/RT}$ value.
0%
Catalyst increases k of reaction.

Explanation

To increase the rate of reaction catalyst does
(i) Decrease

$E_a$
(ii) Decrease

$E_a/RT$
(iii) Increases

$-E_a/RT$
(iv) Increase

$e^{-E_a/RT}$
(v) Increase k.

Which is wrong about a positive catalyst?

0%
It lowers the energy of activation of forward reaction
0%
It lowers the threshold energy level
0%
It lowers the energy of activation of backward reaction
0%
It makes the reaction more exothermic or endothermic

Explanation

Catalyst cannot make the reaction more exothermic or endothermic.

$\Delta H = E_{a(f)} - E_{a(b)}$ is fixed for a reaction since catalyst change the activation by equal magnitude in both forward and backward reaction.

Identify an emulsion from among the following.

0%
Ruby
0%
Sponge
0%
Milk
0%
Jellies

Which of the following cannot be used to prepare an emulsion of benzene in water?

0%
$CH_3-(CH_2)_9-(SO_3)^-Na^+$
0%
$CH_3-(CH_2)_9-COO^-Na^+$
0%
$CH_3-(CH_2)_9-NH_3^+Cl^-$
0%
$(CH_3)_4 \overset{+}{N}Br^-$

Explanation

Surface active agents are used as emulsifiers. The surface-active agents must have a hydrophobic chain with carbon atoms

$\geq$ 5. So, D option is correct.

Which of the following statements about physical adsorption is not correct?

0%
It is usually monolayer
0%
It is reversible in nature
0%
It involves van der Waals interactions between adsorbent and adsorbate
0%
It involves small value of adsorption

Colloidal gold can be prepared by:

0%
bredig's arc method
0%
reduction of $\displaystyle AuCl_{3}$
0%
hydrolysis
0%
peptization

Explanation

Reduction of

$AuCl_3$ :
Colloidal solutions of metals like gold, silver and platinum can be obtained by reduction of their salts.

$2AuCl_3+3SnCl_2\rightarrow 3SnCl_4+2Au$ .

Bredig's arc method:
This method is suitable for preparation of colloidal solutions of metals like gold, silver, platinum etc.
Hydrolysis: for iron
Peptization: for Fe, Ag, Sn and Cd.

Which of the following graphs shows the variation of physical adsorption with temperature?

Consider the following sequence of reactions:

Red Hot Coke

$\ +\$ Steam

$\xrightarrow {\displaystyle \text{Y}{(\text{catalyst})}} X \longrightarrow Z \xrightarrow [\displaystyle \text{cold water}]{\displaystyle \text{pressure}}\$ W(gas)

$\text{Y}$ is:

0%
$V_{2}O_{5}$
0%
$Pt$
0%
$Fe_{2}O_{3}/Cr_{2}O_{3}$
0%
$Ni$

Smoke screens consist of:

0%
fine particles of $TiO_2$ dispersed in air by aeroplanes
0%
fine particles of $AgI$ dispersed in air by aeroplanes
0%
fine particles of $Al_2O_3$ dispersed in air by aeroplanes
0%
none of the above

Explanation

Smoke screen is a cloud of smoke. It consists of fine particles of

$TiO_2$ which are dispersed in air by aeroplanes.

Which of the following statements is/are true for physisorption?

0%
Extent of adsorption increases with increase in pressure.
0%
It needs activation energy.
0%
It can be reversed easily.
0%
It occurs at high temperature.

Which of the following is not correct for heterogeneous catalysis?

0%
The catalyst decreases the energy of activation
0%
The surface of catalyst plays an important role
0%
The catalyst actually forms a compound with reactants
0%
There is no change in the energy of activation

Which statement is incorrect?

0%
Catalytic decomposition of $KClO_3$ by $MnO_2$ is an example of homogeneous catalysis.
0%
Only the surface atoms in an adsorbent play an active role in adsorption.
0%
The rate of adsorption is very rapid in the beginning but decreases gradually until equilibrium is attained.
0%
Absorption proceeds at a steady rate.

Explanation

Catalytic decomposition of

$KClO_3$ by

$MnO_2$ is an example of heterogeneous catalysis.

$2KClO_3(s) \xrightarrow {MnO_2 (s)} 2KCl(s)+3O_2(g)$
It contains two phase solid and gas.
Homogeneous catalysis involves only one phase whereas heterogeneous catalysis involves more than one phase.

How many

$N{H}_{3}$ molecules are adsorbed on charcoal surface?

0%
$1.4\times {10}^{22}$
0%
$1.4\times {10}^{23}$
0%
$1.4\times {10}^{21}$
0%
$1.4\times {10}^{20}$

Explanation

Surface area occupied by one

$N{H}_{3}$ molecule

$=\pi{r}^{2}$ .

$=3.14\times {(\displaystyle\frac{0.3\times {10}^{-9}}{2})}^{2} =7.065\times {10}^{-20}\ {m}^{2}$

Total surface area

$= {10}^{3}\ {m}^{2}$ .

Number of

$N{H}_{3}$ molecules adsorbed

$= \displaystyle\frac{{10}^{3}}{7.065\times {10}^{-20}}$

$= 1.415\times {10}^{22}$ molecules.

Which characteristic is not correct for physical adsorption?

0%
Adsorption is spontaneous.
0%
Both enthalpy and entropy change of adsorption are negative.
0%
Adsorption on solid is reversible.
0%
Adsorption increases with increase in temperature.

Select the wrong statement.

0%
The reducing power of $H_2$ is increased if it is absorbed on $Pd$ .
0%
Thermite welding does not require a catalyst.
0%
The decomposition of $H_2O_2(l)$ in presence of $Hg(l)$ is homogeneous catalysis.
0%
The chemical adsorption requires activation energy.

Explanation

$\begin{array}{l}\text{ The decomposition of }\mathrm{H}_{2}\mathrm{O}_{2}(\mathrm{l})\text{ is heterogenous }\text{catalysis because }\mathrm{O}_{2}(\mathrm{~g})\text{ is formed in }\\\text{ the chemical reaction which is in gaseous }\text{ state .}\end{array}$

Methylene blue is adsorbed on activated charcoal at

$25^oC$ from its aqueous solution. For this process, select the correct statement.

0%
The adsorption requires activation at $25^oC$
0%
The adsorption is accompanied by a decrease in enthalpy
0%
The adsorption increases with increase of temperature
0%
The adsorption is irreversible

Explanation

Methylene blue is adsorbed on activated charcoal at

$25^oC$ from its aqueous solution. For this process, the adsorption is accompanied by a decrease in enthalpy. Adsorption is physisorption and hence, the energy of activation is not required.

Physisorption is exothermic and reversible and it decreases with an increase in temperature. The process, which does not involve any surface reaction, is achieved quickly at low temperatures.

Which statements are correct?
Addition of

$AgNO_3$ drop by drop in excess of

$KI$ leads to the formation of

$+$ vely charged sol of

$AgI$ .
Adsorption characteristic of an adsorbent may be increased by making it in finely divided form.
Catalyst help in attaining the equilibrium earlier but does not influence equilibrium constant in reversible reactions.
Decolorisation of raw sugar is based on absorption.
Gas masks containing animal charcoal acts on the basis of selective adsorption.
The micelle formation of soaps depends upon hydrophobic and hydrophilic entities present in soap.

0%
$2,\: 3,\: 5,\: 6$
0%
$2,\: 3,\: 4,\: 5$
0%
$1,\: 2,\: 5,\: 6$
0%
$1, \:2, \:3, \:4$

Explanation

1. Addition of

$AgNO_3$ in

$KI\:(excess)$ gives

$-$ ve sol due to adsorption of

$I^-$ on

$AgI$ , i.e., preferential adsorption.

2. An increase in surface area of adsorbent increases the adsorption rate.

3. The forward reaction is catalysed and takes place fastly whereas the increase in products concentration makes reverse reaction fast and thus, equilibrium is attained earlier.

4. Animal charcoal adsorbs colouring matter of sugar.

$CH_4$ and

$CO$ are preferentially adsorbed on animal charcoal

6. The micelle formation of soaps depends upon hydrophobic and hydrophlilic entities present in soap and it is reason why soaps forms micelles.

Mention catalyst in the following reaction:

$CS_2+3Cl_2 \rightarrow CCl_4 + S_2Cl_2$

0%
$MnO_2$
0%
$FeCl_3$
0%
$PdCl_2$
0%
$ZnO/CuO$

Explanation

Carbon disulfide reacts with chlorine to produce carbon tetrachloride and disulfur dichloride.

In this reaction, the catalyst could be 1) Manganese (ll) chloride -

$MnCl_2$

2) Aluminium Chloride -

$AlCl_3$

3) Iron(lll) Chloride -

$FeCl_3$

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

$^{-1}$ .

0%
$1\times 10^{-19}m^{2}$
0%
$1.2\times 10^{-19}m^{2}$
0%
$2.5\times 10^{-21}m^{2}$
0%
None of these

Explanation

$400mL$ 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

$\text{B}$

Which of the following statements is incorrect regarding physisorption?

0%
It occurs because of van der Waals forces
0%
Liquefiable gases are adsorbed more easily
0%
Enthalpy of adsorption $(\Delta H_{adsorption})$ is high and negative
0%
Enthalpy of adsorption $(\Delta H_{adsorption})$ is low and positive

Explanation

For physisorption, the enthalpy of adsorption

$(\Delta H_{adsorption})$ is low and negative.
Its value is -20 to -40 kJ/mol.

Note:
Physisorption occurs because of the presence of van der Waals forces between adsorbent and adsorbate molecules.

Liquefiable gases are adsorbed more easily as they have strong intermolecular attractive forces.

Hence, the correct answers are

$\text{C}$ and

$\text{D}$

When dilute aqueous solution of

$AgNO_3$ (excess) is added to

$KI$ solution, positively charged sol particles of

$AgI$ are formed due to adsorption of ion :

0%
$K^+$
0%
$Ag^+$
0%
$I^-$
0%
$NO_3^-$

Explanation

Solution particles possess the tendency to adsorb preferentially the common ion present in solution as follows:

$AgI+Ag^+ \rightarrow Ag_2I^+$

So, option B is correct.

Select incorrect statement(s):

0%
Surface active agent like soaps and synthetic detergents are micelles.
0%
Soaps are emulsifying agents.
0%
$C_{17}H_{35}$ (hydrocarbon part) and COO (carboxylate part) of stearate ion $(C_{17}H_{35}COO^-)$ are both hydrophobic.
0%
The hydrophobic part accumulates inside of the micelle when soap is mixed in water.

Explanation

Option (C) is incorrect because

$C_{17}H_{35}$ is hydrophobic, whereas

$COO^-$ of stearate ion (

${ C }_{ 17 }{ H }_{ 35 }{ COO }^{ - }$ ) is hydrophilic. This owes to the polarity of ions and hydrocarbons.

Select the correct statements about preparation of colloidal solution.

0%
Various substances such as mercury, oil, sulphur etc. are passed in colloidal state by using ultrasonic vibrations.
0%
A precipitate of $BaSO_4$ can be passed in colloidal state by continuous washing with water.
0%
A solution of indigo in suitable solvent is added drop by drop to water forms a colloidal solution of indigo in water.
0%
Peptisation is a characteristic property of colloidal state.

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

0%
dispersion medium is coloured
0%
dispersed phase is coloured
0%
both are coloured
0%
none is coloured

Preparation of Lyophobic sols by chemical method involves :

0%
double decomposition
0%
oxidation and reduction
0%
hydrolysis
0%
all of these

Explanation

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.

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)?

0%
30
0%
40
0%
50
0%
60

Explanation

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)

$\displaystyle =\dfrac {7.227 \times 10^{20}}{2.4\times 10^{13}}=3 \times 10^{7}$

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

0%
$NH_4OH$ , $RCHO$
0%
$RNH_3OH$ , $RCOOH$ or $RSO_3H$
0%
$RNH_3Cl$ , $RSO_4H$
0%
None of these

Explanation

$\begin{array}{l}\text{ Anion and cation resins are used in ion exchange }\\\text{ process. Positive charge is anion resins and }\\\text{ negative charge is cation resins. }\\\text{ Cation exchange resin is cross linked polymer }\\\text{ with negative ion and anion exchange resin is }\\\text{ vice-versa }\end{array}$
1993902_397344_ans_32e2871a74f14a9aab1455b87302d464.png

1993902_397344_ans_32e2871a74f14a9aab1455b87302d464.png

$3$ g of activated charcoal was added to

$50$ mL of the 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 :

0%
$18$ mg
0%
$36$ mg
0%
$42$ mg
0%
$54$ mg

Explanation

$50$ mL of

$0.06\ N$ acetic acid solution contains

$0.06\ mol/L$

$\times \dfrac {50\: mL}{1000\: mL/L} \times 60 \: g/mol = 0.180 \ g$ of acetic acid.

After filtration,

$50$ mL of

$0.042\ N$ acetic acid solution contains

$0.042 \: mol/L \times \dfrac {50 \ mL}{1000 \ mL/L}= 0.126 \ g$ of acetic acid.

Hence, the amount of acetic acid adsorbed on

$3\ g$ of activated charcoal is

$0.180 \: g - 0.126 \: g = 0.054 \: g$ . This is equal to

$54\ mg$ .

Hence, the amount of acetic acid adsorbed on

$1\ g$ of activated charcoal is

$\dfrac {54 \: mg \times 1 \: g}{3 \: g}=18 \: mg$ .

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

0%
P-3, Q-4, R-1, S-2
0%
P-2, Q-3, R-1, S-4
0%
P-3, Q-1, R-4, S-2
0%
P-4, Q-3, R-1, S-2

Explanation

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.

The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of

$KCl, CH_{3}OH$ and

$CH_{3}(CH_{2})_{11}OSO_{3}^{-} Na^{+}$ at room temperature.

The correct assignment of the sketches is:

0%
$I: KCl; II: CH_{3}OH; III: CH_{3}(CH_{2})_{11}OSO^{-}_{3}Na^{+}$
0%
$I: CH_{3}(CH_{2})_{11}OSO^{-}_{3}Na^{+} II: CH_{3}OH; III: KCl$
0%
$I: KCl; II: CH_{3}(CH_{2})_{11}OSO^{-}_{3}Na^{+}; III: CH_{3}OH$
0%
$I: CH_{3}OH; II: KCl; III: CH_{3}(CH_{2})_{11}OSO^{-}_{3}Na^{+}$

Explanation

Impurities affect surface tension appreciably. It is observed that impurities which tend to concentrate on the surface of liquids, compared to its bulk lower the surface tension.

Substances like detergents and soaps

$(CH_3(CH_2)_{11}SO_3^-Na^+)$ decrease the surface tension sharply.

Those like alcohol

$(eg. CH_3OH, C_2H_5OH)$ lower the surface tension slightly. This can also be related to the fact that

$CH_3OH$ has a smaller dielectric constant. The dielectric constant is directly proportional to surface tension. So, on adding

$CH_3OH$ in water, overall dielectric constant decreases and surface tension decreases.

Inorganic impurities present in the bulk of a liquid such as KCl tend to increase the surface tension of water.

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:

0%
$P$ and $Q$
0%
$Q$ and $R$
0%
$P, Q$ and $S$
0%
$P, R$ and $S$

Explanation

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.

Foam is formed when_______

0%
A solid dispersed in a liquid
0%
A liquid dispersed in a liquid
0%
A gas dispersed in a liquid
0%
A liquid dispersed in a gas

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

0%
sodium aluminium silicate is used in the softening of hard water
0%
potash alum is used in shaving rounds and as a styptic in medicine
0%
artificial rain is caused by throwing electrified sand on the clouds from an aeroplane
0%
deltas are formed at a place when the river pours its water into the sea

Physisorption is :

0%
It is highly specific in nature
0%
It is reverisible
0%
It is caused by chemical bond formation
0%
All of the above

Which of the following option is correct for colloids?

0%
Blue color of the sky is due to the scattering of light by colloidal dust particles in air
0%
Sea water looks blue due to scattering of light by the colloidal impurities present in sea water.
0%
Blood is a colloidal solution
0%
All of the above

Smoke particles are actually electrically charged colloidal particles of :

0%
carbon in air
0%
water droplets in air
0%
sulphur in air
0%
titanium oxide in air

The colloidal solution of

$X$ in gelatin is applied in glass plates to form sensitive plates in photography. Identify

$X$ :

0%
Silver chloride
0%
Silver bromide
0%
Sodium chloride
0%
Sodium bromide

Explanation

$AgBr$ (Silver bromide) is used in photography. The colloidal solution of it in gelatin is applied in glass plates to form sensitive plates.

Which of the following is absorbed efficiently in colloidal forms?

0%
Colloidal sulphur
0%
Colloidal gold
0%
Colloidal copper
0%
Colloidal mercury

Soap solution is colloidal in nature and remove the dust particles by which of the following process?

0%
Emulsification
0%
Adsorbtion
0%
Stripping
0%
Distillation

CBSE Questions for Class 12 Engineering Chemistry Surface Chemistry Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Surface Chemistry Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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