MCQ Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 10

An element is oxidised by fluorine and not by chlorine. Identify the element.

0%
Sodium
0%
Aluminium
0%
Oxygen
0%
Sulphur

Explanation

$\begin{array}{l}\text{ Flourine is more Electronegative element than Oxygen and Oxygen is more Electronegative }\\\text{ than Chlorine. }\\\text{ Hence, Flourine can oxidise Oxygen whereas Chlorine reduced Oxygen. }\\\text{ Hence option }(C)\text{ is correct. }\end{array}$

When plants and animals decay, the organic nitrogen is converted into inorganic nitrogen. The inorganic nitrogen is in the form of:

0%
ammonia
0%
elements of nitrogen
0%
nitrates
0%
nitrides

Explanation

When plants and animals decay, the organic nitrogen is converted into inorganic nitrogen. The inorganic nitrogen is in the form of ammonia

$\displaystyle NH_3$ .

The gas produced from thermal decomposition of

$Pb(NO_3)_2$ is:

0%
oxygen
0%
nitric oxide
0%
ammonia
0%
nitrogen

The most abundant element in the earths crust by weight is:

0%
$Si$
0%
$Al$
0%
$O$
0%
$Fe$

The nitride salt of

$Ca$ when treated with

$H_{2}O$ gives:

0%
$N_{2}$
0%
$CaO$
0%
$CaH_{2}$
0%
$NH_{3}$

Explanation

When calcium nitride reacts with di nitrogen it forms ammonia and calcium hydroxide.

$Ca_3N_2 + 6 H_2O\rightarrow 3 Ca(OH)_2 + 2 NH_3$ .

Which of the following noble gases has the highest positive electron gain enthalpy value?

0%
Helium
0%
Krypton
0%
Argon
0%
Neon
0%
Xenon

Explanation

Noble gas	Electron gain enthalpy $(kJ\ mol^{-1})$
$He$	$48$
$Ne$	$116$
$Ar$	$96$
$Kr$	$96$
$Xe$	$77$

The catalyst used in the manufacture of nitric acid by Ostwald's process is:

0%
$V_2O_5$
0%
$Pt$
0%
$Mo$
0%
$Fe$

Which product will be obtained in the following reaction?

$P_4(s) + 3NaOH(ag) + 3H_2O(l) \rightarrow$

0%
$PH_3(g) + 3NaH_2P0_2 (aq)$
0%
$PH_3 (g) + 3Na_2 HPO_2 (aq)$
0%
$2PH_3(g) 3Na_2 HPO_2 (aq)$
0%
$2PH_3(g) + 3NaH_2 PO_2 (aq)$

Explanation

$P_4(s) + 3NaOH(ag) + 3H_2O(l) \rightarrow \underset{Phosphine}{PH_3(g)}+\underset{Sodium \, hypophosphite}{3NaH_2PO_2(aq)}$
This reaction is used for the preparation of phosphine gas in the laboratory.

In the Haber process of synthesis of

$NH_3$ , which of the following statement is correct?

0%
$Mo$ acts as a catalyst and $Fe$ as a promoter
0%
$Fe$ acts as a catalyst and $Mo$ as a promoter
0%
$Fe$ acts as inhibitor and $Mo$ as a catalyst
0%
$Fe$ acts as promoter and $Mo$ as autocatalyst

Explanation

Preparation of Ammonia:

$N_2 + 3H_2\rightarrow 2NH_3$

At high pressure about 200-250 atm, and a moderate temperature of $750K$ , the reaction takes place. Removal of ammonia is done by liquefaction, In the process, iron is used as a catalyst and molybdenum as a promoter.

Which one of the following statements is incorrect?

0%
All halogens have weaker X-X bond than X- X' bond, in interhalogens (except F-F bond in fluorine)
0%
Radius ratio between iodine and fluorine (among halogens) is maximum
0%
Interhalogen compounds are more reactive than halogen compounds
0%
Among interhalogen compounds, maximum number of atoms are present in iodine fluoride

Explanation

All interhalogens have weaker

$X-X'$ bonds and not

$X-X$ bonds but F is an exception due to high electronegativity and small size.

Hence, option (A) is incorrect.

Hydrolysis of

$N{ Cl }_{ 3 }$ gives

$N{ H }_{ 3 }$ and

$X$ , which of the following is

$X$ ?

0%
$HCl{ O }_{ 4 }$
0%
$HCl{ O }_{ 3 }$
0%
$HOCl$
0%
$HCl{ O }_{ 2 }$

Explanation

Hydrolysis of

$N{ Cl }_{ 3 }$ gives

$N{ H }_{ 3 }$ or

$N{ H }_{ 4 }OH$ and

$HClO$ as

$N{ Cl }_{ 3 }+4{ H }_{ 2 }O\longrightarrow N{ H }_{ 4 }OH+3HOCl$

The maximum covalency of nitrogen is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Which of the following phosphorus is the most reactive?

0%
Red phosphorus
0%
White phosphorus
0%
Scarlet phosphorus
0%
Violet phosphorus

Noble gas are adsorbed by:

0%
anhydrous calcium chloride
0%
ferric hydroxide
0%
concentrated ${ H }_{ 2 }S{ O }_{ 4 }$
0%
activated coconut charcoal

Platinum is used as a catalyst in:

0%
oxidation of ammonia to form nitric acid
0%
hydrogenation of oils
0%
production of synthetic rubber
0%
synthesis of methanol

Explanation

In the Ostwald process, ammonia is converted to nitric acid in a two step process. It is oxidised with oxygen in the presence of platinum catalyst. The overall reaction is as follows:

$4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O$

In pyrophosphoric acid,

${ H }_{ 4 }{ P }_{ 2 }{ O }_{ 7 }$ , number of

$\sigma -$ and

$d\pi -p\pi$ bonds respectively are

0%
$8$ and $2$
0%
$6$ and $2$
0%
$12$ and zero
0%
$12$ and $2$

Explanation

$O=\overset{\,\,\,\,\,\,O-H}{\overset{|}{\underset{\,\,\,\,\,O-H}{\underset{|}{P}}}}-O-\overset{\,\,\,\,\,\,O-H}{\overset{|}{\underset{\,\,\,\,\,O-H}{\underset{|}{P}}}}=O$

The two double bonds contributes 2

$\sigma$ and 2

$d\pi-p\pi$ .
There are 4 O-H groups = 4

$\sigma$ bonds
There are 6 P-O type of bonds.

Total no. of

$\sigma$ bonds = 2+ 4+ 6 =12
Total no. of

$d\pi-p\pi$ bonds= 2

Hence, the correct option is D.

In the Haber process for the industrial manufacturing of ammonia involving the reaction

${ N }_{ 2 }(g)+3{ H }_{ 2 }(g)\rightleftharpoons 2{ NH }_{ 3 }(g)$ at

$200atm$ pressure in the presence of a catalyst, a temperature of about

${ 500 }^{ o }C$ is used. This is considered as optimum temperature for the process because:

0%
yield is maximum at this temperature
0%
catalyst is active only at this temperature
0%
energy needed for the reaction is easily obtained at this temperature
0%
rate of the catalytic reaction is fast enough while the yield is also appreciable for this exothermic reaction at this temperature

Explanation

In order to yield the maximum yield of ammonia, the temperature is lowered though

$500^oC$ is not a low temperature. This temperature

$(500^oC)$ is a compromise temperature to produce reasonably a high proportion of ammonia in the equilibrium mixture in a very short time.

Which can do glass etching?

0%
$HI{ O }_{ 4 }$
0%
$Si{ F }_{ 4 }$
0%
$HF$
0%
$H{ NO }_{ 3 }$

Explanation

Glass is a mixture of sodium and calcium silicate. It reacts with

$HF$ forming sodium and calcium fluorosilicates respectively.

${ Na }_{ 2 }Si{ O }_{ 3 }+6HF\longrightarrow { Na }_{ 2 }Si{ F }_{ 6 }+3{ H }_{ 2 }O$

$CaSi{ O }_{ 3 }+6HF\longrightarrow CaSi{ F }_{ 6 }+3{ H }_{ 2 }O$
The etching of glass is based on these reactions.

The oxidation state of nitrogen in dinitrogen trioxide is:

0%
$+1$
0%
$+2$
0%
$+3$
0%
$+4$

Explanation

Dinitrogen trioxide is

$N_2O_3$ .

The oxidation state of nitrogen in

$N_2O_3$ is:

$2\times x+3\times(-2)=0$

$x=+3$

Therefore, option

$C$ is correct.

Which compound of nitrogen is formed when

$CaNH$ reacts with hot water?

0%
$NH_{3}$
0%
$N_{2}O$
0%
$NH_{2}NH_{2}$
0%
$NO_{2}$

Explanation

When $CaNH$ reacts with hot water it forms $NH_3,Ca(OH)_2$ .

$CaNH+2H_2O\rightarrow Ca(OH)_2+NH_3$ .

Henceoption is correct.

The heat of atomisation of

${ P }_{ 2 }{ H }_{ 4 } (g)$ and

${ P }{ H }_{ 3 } (g)$ are 355 kcal/mol and 228 kcal/mol respectively. The energy of

$P-P$ bond is:

0%
$102\ kcal\ { mol }^{ -1 }$
0%
$31\ kcal\ { mol }^{ -1 }$
0%
$26\ kcal\ { mol }^{ -1 }$
0%
$204\ kcal\ { mol }^{ -1 }$

Consider the equilibrium

$NH_{4}Cl (s)\rightleftharpoons NH_{3}(g) + HCl(g)$ . An inert gas is added to the system at constant volume and temperature. The correct statement(s) is/ are:

0%
the partial pressures of $NH_{3}$ and $HCl$ in the system will increase
0%
the partial pressures of $NH_{3}$ and $HCl$ in the system will remain the same
0%
the partial pressures of $NH_{3}$ and $HCl$ in the system will decrease
0%
the entropy of the system will increase

Explanation

$NH_4$

$+$

$Cl(s)$

$\rightleftharpoons$

$NH_3 (g)$

$+$

$HCl$ .

Noble gas are inert hence adding inert gas doesn't effect the partial pressure.

This is because noble gas gets added to both reactant and product.

$X (g)$

$+$

$NH_4Cl (S)$

$\rightleftharpoons$

$NH_3(g)$

$+$

$HCl(g)$

$+$

$X(g)$

Where X=Noble gas

Hence the partical presence of

$NH_3$ and

$HCl$ will remain same.

The amount of gas in the system will increase after adding inert gas.

Since gases has high entropy than liquid and solid, the entropy of the system will increase.

Which gas is produced when calcium nitride

$(Ca_{3}N_{2})$ is hydrolyzed by water?

0%
$N_{2}$
0%
$NH_{3}$
0%
$H_{2}$
0%
$O_{2}$

Explanation

Ammonia gas is produced when calcium nitride

$(Ca_{3}N_{2})$ is hydrolyzed by water.

$Ca_3N_2+6H_2O \to 3Ca(OH)_2+2NH_3$

Formation of ammonia will be favorable in the following cases(s):

$N_2(g)\,+\,3H_2(g)\rightarrow2NH_3(g)$

$\Delta H\,<\,0$

0%
Addition of inert gas at constant volume
0%
Decreasing the volume of the container
0%
Addition of inert gas at constant pressure
0%
Increasing the pressure

Explanation

In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. The optimum conditions for the production of ammonia are a pressure of 200 × 105 Pa (about 200 atm), a temperature of ~ 700 K and the use of a catalyst such as iron oxide with small amounts of $K_2O$ and $Al_2O_3$ to increase the rate of attainment of equilibrium.
Addition of an inert gas at constant volume: When an inert gas is added to the system in equilibrium at constant volume, the total pressure will increase. But the concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change.
Hence, option A is correct answer as addition of ideal gas increases pressure and thus favours formation of Ammonia.

The electronic configuration of few elements is given below. Mark the statement which is not correct about these elements.
(i)

$1s^{2} 2s^{2} 2p^{6} 3s^{1}$
(ii)

$1s^{2} 2s^{2} 2p^{5}$
(iii)

$1s^{2}2s^{2}2p^{6}$
(iv)

$1s^{2}2s^{2}2p^{3}$

0%
(i) is an alkali metal.
0%
(iii) is a noble metal.
0%
(i) and (ii) form ionic compounds.
0%
(iv) has high ionisation enthalpy.

The electronic configuration of four atoms are given in brackets:
L : (

$1s^2\,2s^2\,2p^1$ ); M : (

$1s^2\,2s^2\,2p^5$ );

Q : (

$1s^2\,2s^2\,2p^6\,3s^1$ ); R : (

$1s^2\,2s^2\,2p^2$ ).

The element that would most readily form a diatomic molecule is :

0%
Q
0%
M
0%
R
0%
L

Which product will obtained in the following reaction?
Reaction:

${ P }_{ { 4 }_{ (a) } }+3Na{ OH }_{ (aq) }+3{ H }_{ 2 }{ O }_{ (l) }\quad \longrightarrow$

0%
${ PH }_{ { 3 }_{ (a) } }+3{ Na }_{ 2 }H{ PO }_{ { 2 }_{ (aq) } }\quad$
0%
${ PH }_{ { 3 }_{ (a) } }+3Na{ H }_{ 2 }{ PO }_{ { 2 }_{ (aq) } }\quad$
0%
$2{ PH }_{ { 3 }_{ (a) } }+3{ Na }_{ 2 }H{ PO }_{ { 2 }_{ (aq) } }$
0%
$2{ PH }_{ { 3 }_{ (a) } }+3Na{ H }_{ 2 }{ PO }_{ { 2 }_{ (aq) } }\quad$

Explanation

Phosphine is formed in the reaction,

$P_4+3NaOH+3H_2O\longrightarrow PH_3+3NaH_2PO_2$
Sodium hydrophosphite is also one of the products of the reaction.

Which among the group-

$15$ elements does not exists as tetra atomic molecule?

0%
Nitrogen
0%
Phosphorus
0%
Arsenic
0%
Antimony

Explanation

Nitrogen does not exist as tetra atomic molecule. It exists as diatomic molecule

$N_2$ that can be represented as

$N \equiv N$ .

Which of the following is used to prepare

$Cl_{2}$ gas at room temperature from concentrated

$HCl$ ?

0%
$MnO_{2}$
0%
$H_{2}S$
0%
$KMnO_{4}$
0%
$Cr_{2}O_{3}$

Explanation

Manganese dioxide( $MnO_2$ ) is heated with concentrated $HCl$ to form $Cl_2$

$MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O$

$HCl$ in action with potassium permanganate gives $C_2$

$2KMnO_4+16HCl\rightarrow 2KCl+2MnCl_2+8H_2O+5Cl_2$ .

Hence option A and C is correct.

Xenon has closed shell configuration but is known to give compounds with fluorine because:

0%
Xe atom has large size and lower ionisation potential as compared to other noble gases
0%
Xe has unpaired electrons which can form covalent bonds
0%
Xe has highest boiling point hence it can form compounds with fluorine
0%
fluorine is the smallest element hence it can react with all noble gases

Explanation

$Xe$ is inert though it gives compounds with Fluorine due to its large size and lower ionization potential as compared to other noble gases. Also, it is more reactive than other noble gases.

Which of the following factors would favour the formation of ammonia?

0%
High pressure
0%
Low temperature
0%
High volume
0%
Low pressure

Explanation

According to Le Chatelier's principle, high pressure would favour the formation of

$NH_{3}$

$N_{2}$ (g) + 3

$H_2$ {g)

$\leftrightharpoons$

$NH_{3}$ (g)
The optimum conditions for the production of

$NH_{3}$ are pressure of 200 x 10

$^{5}$ Pa (about 200 atm) and temperature of -700 K.

In the given reaction, the number of moles of reactant is more than the product and according to the ideal gas equation, the pressure is inversely proportional to the number of moles.

Thus, by increasing pressure i.e. at high pressure, the reaction will proceed in a forward direction towards less number of moles. Thus the formation of the ammonia increases.

A cold, green flame can be made by passing carbon dioxide over warm.

0%
Grey tin
0%
White phosphorous
0%
Bronze
0%
Green candles

The structure of white phosphorus is:

0%
square planar
0%
pyramidal
0%
tetrahedral
0%
trigonal planar

Explanation

Phosphorus or s tetraphosphorus ( $P_4$ ) exists as molecules made up of four atoms in a tetrahedral structure. The molecule is described as consisting of six single P–P bonds.

Hence option C is correct.

The halogen that is most easily reduced is:

0%
$F_{2}$
0%
$Cl_{2}$
0%
$Br_{2}$
0%
$I_{2}$

Explanation

$F_2$ is easily reduced as

$F^{-1}$ is the strongest oxidising agent. Oxidising power can be explained by the fact that the ease with which a halogen

$[X]$ forms the halide ion

$H^-$ decreases down the group.

Choose the correct statements from the following?

0%
Rhombic sulphur is blue in colour.
0%
Rhombic sulphur is soluble in water but insoluble in organic solvents.
0%
Rhombic and monoclinic sulphur have $S_{6}$ molecules.
0%
In Cyclo- $S_6$ molecule, the ring adopts chair form.

Explanation

Rhombic sulphur is yellow in colour.

Rhombic sulphur cannot be dissolved in water but can be dissolved in benzene, ether, alcohol, etc.

Both rhombic and monoclinic Sulphur have the molecular formula

$S_8$ .

In cyclo -

$S_6$ molecule, the ring adopts chair form as shown in the figure.

Option D is correct.

1157047_935421_ans_5e15456b0d6348d9908d59261ab12170.png

Which of the following compounds will not give ammonia on heating?

0%
$(NH_{4})_{2}SO_{4}$
0%
$(NH_{4})_{2}CO_{3}$
0%
$NH_{4}NO_{2}$
0%
$NH_{4}Cl$

Explanation

$NH_4NO_2$ will not give ammonia on heating.It gives $NO_2H_2O$

$NH_4NO_2(s)\rightarrow N_2(g) + 2H_2O(g)$

Hence option C is correct.

Which of the following gases cannot be kept in a glass bottle because it chemically reacts with glass?

0%
Sulphur
0%
Chlorine
0%
Fluorine
0%
Methane

Which of the following statements is not correct?

0%
All the oxides of halogens are powerful oxidants.
0%
The compounds of oxygen and fluorine are not called oxides but fluorides.
0%
Oxygen fluorides form oxoacids.
0%
In oxyhalides, bonds are mainly covalent due to small difference in electronegativity of oxygen and halogens.

Explanation

Oxygen florides forms $HF,O_2$ on hydrolysis.

$OF_2 + H_2O\rightarrow O_2 + 2 HF$

$2O_2F_2 + 2H2O\rightarrow O_2 + 4HF$

so oxygen florides do not form forms hypohalous acids.

So the given statement oxygen florides forms oxoacids is wrong.

Hence option C is correct.

Which compound is prepared by the following reaction?
Xe +

$F_{2}$ (2:1 volume ratio)

$\overset{Ni}\rightarrow$ at 673 K

0%
$XeF_{4}$
0%
$XeF_{2}$
0%
$XeF_{6}$
0%
None of these

Explanation

$Xe+F_2\overset {673K}{\underset {Ni}{\longrightarrow}} XeF_2\quad (1:3$

$Ratio)$

$Xe+2F_2\overset {873K}{\underset {Ni}{\longrightarrow}} XeF_4\quad (1:5$

$Ratio)$

$Xe+3F_2\overset {673K}{\underset {Ni}{\longrightarrow}} XeF_6\quad (1:20$

$Ratio)$

Oxygen can be prepared by heating ___________.

0%
potassium permanganate
0%
methane
0%
carbon
0%
none of these

Explanation

Oxygen can be prepared by heating potassium permanganate. On heating, potassium permanganate splits into potassium manganate and oxygen.

$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

In the preparation of

$HNO_{3}$ , we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of

$NH_{3}$ will be :

0%
2
0%
3
0%
4
0%
6

Explanation

In the preparation of

$HNO_{3}$ , we get NO gas by catalytic oxidation of ammonia.In the below equation 4 moles of

$NH_3$ produced 4 moles of

$NO$ .So the moles of

$NO$ produced by the oxidation of two moles of

$NH_{3}$ will be2 moles.

$4NH_3+5O_2\rightarrow 4NO+6H_2O$ .

Hence option A is correct.

Which of the following statements is wrong?

0%
Single N-N bond is stronger than the single P-P bond.
0%
$PH_{3}$ can act as a ligand in the formation of coordination compound with transition elements.
0%
$NO_{2}$ is paramagnetic in nature.
0%
Covalency of nitrogen in $N_{2}O_{5}$ is four.

Explanation

N-N sigma bond (single bod) is weaker than P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogenrepel each other making the single bon between $N-N$ weaker than P-P sigma bond.In $P-P$ bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.

So statement A is wrong.

Hence option A is correct.

Which of the following statements is not correct about

$XeF_{2}$ ?

0%
It can be obtained by direct reaction between $F_{2}$ and Xe at high pressure.
0%
$XeF_{2}$ undergoes alkaline hydrolysis to give $O_{2}$ and Xe.
0%
$XeF_{2}$ is a powerful reducing agent.
0%
$XeF_{2}$ contains two bond pairs and three lone pairs.

Explanation

$XeF_2$ is obtained by direct reaction of

$Xe$ and

$F_2$ at high pressure. It undergoes alkali hydrolysis to produce

$Xe$ and

$O_2$ .

$XeF_2$ is a powerful reducing agent where

$Xe^{+2}$ can change to

$+6$ state.

The product which is not formed on the heating of potassium permanganate is ________.

0%
$K_2MnO_4$
0%
$MnO_2$
0%
$O_2$
0%
$MnO_4$

Explanation

On heating potassium parmanganate we do not get

$MnO_4$ . All the other products are formed on heating potassium permanganate.

$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

Which factor is most responsible for the increase in boiling points of noble gases from

$He$ to

$Xe$ ?

0%
Decrease in $I.E.$
0%
Monoatomic nature
0%
Decrease in polarisability
0%
Increase in polarisability

Explanation

Polarisability increases with increase in size. So, from

$He$ to

$Xe$ as size increases, polarisability increases.

Which of the following substance will have highest boiling point?

0%
$He$
0%
$CsF$
0%
${ NH }_{ 3 }$
0%
$CH{ Cl }_{ 3 }$

Explanation

Boiling point

$\times$ Molar mass

$\times$ Ionic character

which is highest for

$CsF$

While preparing oxygen from potassium permanganate, the hard glass test tube must be held steady.

0%
True
0%
False

Which of the molecule does not exist?

0%
$ClF_3$
0%
$ICl_3$
0%
$BrF_3$
0%
$ClI_3$

Explanation

$ClI_3$ does not exist as

$I$ is bigger than

$Cl$ so

$Cl$ cannot have

$3I$ as having

$3I$ on

$Cl$ will become crowded.

Which of the following displaces

$Br_{2}$ from an aquenous solution containing bromide ions?

0%
$Cl^{-}$
0%
$Cl_{2}$
0%
${ I }^{-}_{3}$
0%
$I_{2}$

In Haber's process of manufacturing of ammonia:

${ N }_{ 2 }\left( g \right) +{ 3H }_{ 2 }\left( g \right) \longrightarrow 2{ NH }_{ 3 }\left( g \right) ;{ H }_{ { 25 }^{ 0 }C }^{ o }=-92.2kJ$

Molecule:

${ C }_{ p }{ JK }^{ -1 }{ mol }^{ -1 }$

${ N }_{ 2 }\left( g \right)$ : 29.1

${ H }_{ 2 }\left( g \right)$ : 28.8

${ NH }_{ 3 }\left( g \right)$ : 35.1
If

${ C }_{ p }$ is independent of temperature, then reaction at

${ 100 }^{ 0 }C$ as compared to that of

$25^{ 0 }C$ will be:

0%
more endothermic
0%
less endothermic
0%
more exothermic
0%
less exothermic

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The P-Block Elements Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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