Chemistry NCERT Exemplar Solutions Class 12th Chapter Five: Overview, Questions, Preparation

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Medicines are more effective in the colloidal form because of large surface area and are easily assimilated in this form.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: On adding dispersion medium, emulsions can be diluted to any extent. The dispersed phase forms a separate layer if added in excess.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: After the reaction is over between adsorbed reactants, the process of desorption is important to remove products and further create space for the other reactant molecules to approach the surface and react.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The gaseous molecules diffuse on to the surface of the solid catalyst and get adsorbed. After the required chemical changes the products diffuse away from the surface of the catalyst leaving the surface free for more reactant molecules to get adsorbed and undergo reaction.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: When gaseous molecules come in contact with the surface of a solid catalyst, a weak chemical combination takes place between the surface of the catalyst and the gaseous molecules, which increases the concentration of reactants on the surface. Different molecules adsorbed side by side have better chance to react and form new molecules. This enhances the rate of reaction. Also, adsorption is an exothermic process. The heat released in the process of adsorption is utilised in enhancing the reaction rate.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The rate of an enzyme reaction is maximum at a definite temperature, called the optimum temperature. On either side of the optimum temperature, the enzyme activity decreases. The optimum temperature range for enzymatic activity is 298-310 K. Human body temperature being 310 K is suited for enzyme-catalysed reactions. Therefore, during fever, catalytic activity of the enzyme may get affected.

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iii)

In homogeneous catalysis, the reactant as well as the product both are in same phase and composition is uniform throughout hence the correct option is (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv)

In equilibrium ΔG=0 so ΔG=ΔH-TΔS which means ΔH=TΔS therefore the correct option is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv) 

As gas-gas forms up a homogeneous composition, so the correct option is (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iii)

When both adsorption and absorption occur simultaneously it is known as sorption; hence the correct option is (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (ii)

As the process of adsorption is an exothermic one; physical adsorption occurs readily at low temperature and decreases with an increase in the temperature as equilibrium shifts in the backward direction. This is known as Le-Chatelier's principle, hence the correct option is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (i)

Extent of adsorption depends on the concentration of the solute in a solution. The concentration of adsorbate increases the interaction between adsorbate and adsorbent thus the extent of adsorption also increases; hence the correct option is (i).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  Correct option is (i)

As adsorption is an exothermic process so? H cannot be greater than zero, hence the correct option is (i).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (i)

High temperature is not favourable for physical adsorption since it is an exothermic process, hence the correct option is (i).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (ii)

On increasing the temperature activation energy of the adsorbate molecule increases which convert physical adsorption into chemisorptions hence the correct option is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (i)

Physisorption adsorbent does not show specificity for any particular gas because it is involved van there Waals forces are universal, hence the correct option is (i).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iii)

In absorption, a substance is uniformly distributed, through the body of the solid or liquid, hence the correct option is (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv)

Lesser the value of critical temperature of gases, the lesser will be the force of attraction among molecules and least will be the adsorption hence the answer is option (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (i)

When the reactant and catalyst are in different phase it is known as heterogeneous catalysis hence the answer is (i).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar 

Ans: Correct option is (ii)

There are few substances which at low concentrations behave as normal strong electrolytes, but at higher concentrations exhibit colloidal behavior due to the formation of aggregates. The aggregated particles formed are called micelles. These are also known as associated colloids hence the answer is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (ii).

Tyndall effect is defined as the optical property shown by colloidal particle. Above the critical micelle concentration, a solution of soap behaves as an associated colloid so it shows Tyndall effect hence the answer is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iii).

Lyophilic colloids have unique property of protecting lyophobic colloids. When a lyophilic sol is added to lyophobic solution, the lyophilic particles form a layer around the lyophobic particles and protect the latter from electrolytes. Lyophilic colloids used for this purpose are called protective colloids hence the answer is (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv)

Peptisation is the process in which freshly prepared precipitate can be converted into colloidal solution so the answer is (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (ii)

The rate of coagulation will be faster if the value of oppositely charge electrolyte is high hence the answer is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv).

Solid and liquid together forms up sol. In this case, solid is dispersed phase and liquid is the dispersion medium hence the answer is (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (iv).

Colloidal particles are bigger aggregate than the number of particles in a colloidal solution hence, the values of

colligative are of small order as compared to values shown by true solutions at same concentration so the answer is (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct answer is (ii)

The correct sequence of steps involved in catalysis are:

(i) Adsorption of A and B on surface

(ii) Interaction between A and B to form intermediate

(iii) Starting of desorption from surface

(iv) Complete desorption from the surface

Therefore, the correct answer is (ii)

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is iii.

River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water, the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition with the formation of delta hence the answeris (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct answer is (iv)

Charge on the sol particles can be a result of the following:

Due to electron capture by sol particles during electrodispersion of metals,

Due to preferential adsorption of ions from solution and/or

Due to formulation of electrical double layer.

Hence the correct answer is (iv)

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: Correct option is (ii)

In the figure adsorption of coloured particle from charcoal is shown. Solution of raw sugar is filtered by animal charcoal

and yellowish brown colour of raw sugar is adsorbed and filterate is colourless which gives white colour on crystallization

hence the answer is (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (ii) and (iii)

The formation of micelles takes place only above a particular temperature which is the Kraft temperature (Tk ) and above a particular concentration i.e., the critical micelle concentration (CMC). Upon dilution, these colloids revert back to the individual ions hence the answer is (ii) and (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (i) and (ii)

The action of a catalyst is selective in nature and so a substance which acts as a catalyst in one reaction may fail to catalyse another reaction. They also do not change the enthalpy of reaction hence the answer is (i) and (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (i) and (iii)

Freundlich gave an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbed and pressure at a particular temperature.

$\frac{\mathrm{x}}{\mathrm{p}}$ =kp $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{n}$}\right.$

If $\frac{1}{\mathrm{n}}$  = 0 ; $\frac{\mathrm{x}}{\mathrm{m}}$  = k xtent of adsorption is independent of pressure

When n=0 ; $\frac{\mathrm{x}}{\mathrm{m}}$  = kp

$\frac{\mathrm{x}}{\mathrm{m}}$  vs p is a line

parallel to x-axis

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:   (ii) and (iii)

H₂ molecule on an activated charcoal is adsorbed to a very little extent in comparison to easily liquefiable gases because it has (a) Very weak van there Waals force of attraction (b) Very low critical temperature hence the answer is (ii) and (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  (ii) and (iv)

The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal

solution, because the repulsive forces between charged particles having same charge prevent them from coalescing or

aggregating when they come closer to one another. The Brownian movement has stirring effect which does not permit the particles to settle and thus, is responsible for the stability of sols hence the answer is (ii) and (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (ii) and (iv)

Emulsions can be broken into constituent liquids by heating, freezing and centrifuging hence the answer is (ii) and (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (i) and  (iv)

Negatively charged emulsion can  be  precipitated by oppositely  charged electrolyte. Na+ and K+ from the  electrolyte can  neutralize  the  negatively  charge  emulsion and  precipitate  the  colloid so the answers are (i) and  (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  (iii)  and  (iv)

Sols  directly  formed  by mixing substances  like  gum, gelatin,   starch, rubber, etc., with a  suitable liquid (the  dispersion 

medium) are  called lyophilic sols. They are  also known as  reversible colloid.  These sols  are  very stable  and cannot

coagulate easily hence the answers are (iii)  and  (iv).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  (i) and  (iii)

Lyophilic colloids have a unique property of protecting lyophobic colloids. When a lyophilic sol is added to the lyophobicm sol, the lyophilic particles form a layer around lyophobic particles and thus protect the latter from electrolytes. Lyophilic colloids used for this purpose are called protective colloid so the answers are (i) and  (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  (ii) and (iii).

When electrophoresis, i.e, movement of particles is prevented by some suitable means, it is observed that the dispersion medium begins to move in an electric field. This phenomenon is termed electroosmosis so the answers are (ii) and (iii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans:  (i) and (ii)

Substances which accelerate the rate of a chemical reaction and themselves remain chemically and quantitatively unchanged after the reaction, are known as catalysts they can undergo physical change so the answers are (i) and (ii).

This is a Multiple Choice Type Questions as classified in NCERT Exemplar

Ans: (i) and (iv)

When a chalk stick is dipped in ink, the surface retains the colour of the ink due to adsorption of coloured molecules

while the solvent of the ink goes deeper into the stick due to absorption so the answers are (i) and (iv).

This is a Matching Type Questions as classified in NCERT Exemplar

Ans : (i)- (b); (ii)- (c); (iii)- (d); (iv)- (a)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i)- (c); (ii)- (d); (iii)- (b); (iv)- (a)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i)- (b); (ii)- (c); (iii)- (d); (iv)- (a)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i) - (b); (ii) - (c); (iii) - (d); (iv) - (a)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: (i)- (d); (ii)- (c); (iii)- (a); (iv)- (b)

This is a Matching Type Questions as classified in NCERT Exemplar

Ans: The answers are: (i)- (d) (ii)- (c) (iii)- (a) (iv)- (b)

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Colloidal particles can pass through ordinary filter paper because the pores are too large. However, the pores of filter paper can be reduced in size by impregnating with collodion solution to stop the flow of colloidal particles hence the answer is (iii).

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Colloidal particles being bigger aggregates, the number of particles in a colloidal solution is comparatively small as compared to a true solution. Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of small order as compared to values shown by true solutions at same concentration hence the answer is (i).

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Colloidal particle shows Brownian movement. The Brownian movement has a stirring effect which does not permit the particles to settle and thus, is responsible for the stability of sols so the answer is (v).

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Greater the valence of the flocculating ion added, the greater is its power to cause precipitation. This is known as Hardy schulze rule. In the coagulation of a negative sol, the flocculating

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: Detergents with low CMC are more economical to use as they involve formation of micelle which is used for cleaning of oil and dirt from our cloth. Micelle formation takes place when the concentration of the detergent becomes equal to the CMC so the answer is (i).

This is a Long Answers Type Questions as classified in NCERT Exemplar

Ans: In heterogeneous catalysis reactants are generally in gas phase and catalyst are in solid phase. The activity of a catalyst depends upon the strength of chemisorption to a large extent. The reactants must get adsorbed reasonably strongly on to the catalyst to become active. However, they must not get adsorbed so strongly that they are immobilized and other reactants are left with no space on the catalyst's surface for adsorption. It has been found that for hydrogenation reaction, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity being shown by groups 7-9 elements of the periodic table. Catalyst also directs a reaction to yield a particular product. For example, starting with H₂ and CO, and using different catalysts, we get different products.

This is a Long Answers Type Questions as classified in NCERT Exemplar

Ans:  (i) Separation of inert gases: Due to a difference in the degree of adsorption of gases by charcoal, a mixture of noble gases can be separated by adsorption on coconut, a charcoal at different temperatures.

(ii) Adsorption indicators: Surfaces of certain precipitates such as silver halides have the property of adsorbing some dyes like eosin, fluorescein, and so produce a characteristic colour at end point.

(iii) Chromatographic analysis: It is based on the phenomenon of adsorption and has a number of applications in analytical and industrial fields.

This is a Long Answers Type Questions as classified in NCERT Exemplar

Ans: This method is used for removing gangue from sulphide ores. In this, a suspension of powdered ore is made with water. To which both the collectors and froth stabilisers are added. Collectors for example pine oils, fatty acids, xanthates enhance non-wettability of the mineral particles and froth stabilisers such as cresols, aniline stabilise the froth.

This is a Long Answers Type Questions as classified in NCERT Exemplar

Ans: Catalytic reaction which depends on the pore structure of catalyst and size of the reactant, product molecules is known as shape-selective catalysis. Zeolites are considered as good shape-selective catalysts due to their honeycomb-like structures. They are aluminosilicates with three-dimensional network of silicates wherein some of the silicon atoms are replaced by aluminium atoms giving Al-O-Si frame work. The reactions taking place in zeolites depends upon size and shape of reactant and product molecules, the pores and cavities of the zeolites also. They are found in nature and synthesised for catalysts selectivity. Zeolites are widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: It is important to have clean surface as it facilitates the adsorption of desired species.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Chemisorption involves formation of bond between gaseous molecules/atoms and the solid surface for which high activation energy is required. Thus it is referred to as activated adsorption.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: At lower concentration soap forms a normal electrolytic solution with water. After a certain concentration called critical micelle concentration, colloidal solution is formed.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Gold sol is a lyophobic sol. Addition of gelatin stabilises the gold sol.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Clouds are colloidal in nature and carry charge. Spray of silver iodide, an electrolyte, results in coagulation leading to rain.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Ice creams are emulsions which get stabilised by emulsifying agents like gelatin.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: It is a 4% solution of nitrocellulose in a mixture of alcohol and ether.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The colloidal impurities present in water get coagulated by added alum, thus making water potable.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The charged colloidal particles start moving towards oppositely charged electrodes.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Unbalanced bombardment of the particles of dispersed phase by molecules of dispersion medium causes brownian motion. This stabilises the sol.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Positively charged sol of hydrated ferric oxide is formed and on adding excess of NaCl, negatively charged chloride ions coagulate the positively charged sol of hydrated ferric oxide.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The emulsifying agent forms an interfacial layer between suspended particles and the dispersion medium thereby stabilising the emulsion.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Animal hide is colloidal in nature and has positively charged particles. When it is soaked in tannin which has negatively charged colloidal particles, it results in mutual coagulation taking place.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: In Cottrell precipitator, charged smoke particles are passed through a chamber containing plates with charge opposite to the smoke particles. Smoke particles lose their charge on the plates and get precipitated.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Minimum quantity of an electrolyte required to cause precipitation of a sol is called its coagulating value. Greater the charge on flocculating ion and smaller is the amount of electrolyte required for precipitation, higher is the coagulating power of coagulating ion (Hardy-Schulze rule).

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Moist alum coagulates the blood and so formed blood clot stops bleeding.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The adsorption of positively charged Fe³⁺ ions by the sol of hydrated ferric oxide results in positively charged colloid.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Physisorption involves weak van there Waals forces which weaken with rise in temperature. The chemisorptions involves formation of chemical bond involving activation energy and like any other chemical reaction is favoured by rise in temperature.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Due to excessive dialysis, traces of electrolyte which stabilises the colloids is removed completely, making the colloid unstable. As a result, coagulation takes place.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Eosin is adsorbed on the surface of silver halide precipitate making it coloured.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: Activated charcoal acts as an adsorbent for various poisonous gases present in the coal mines.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: River water is a colloidal solution of clay and sea water contains lot of electrolytes. The point at which river and sea meet is the site for coagulation. Deposition of coagulated clay results in delta formation.

This is a Short Answers Type Questions as classified in NCERT Exemplar

Ans: The process of physisorption for example that of H2 on finely divided nickel, involves weak van there Waals' forces. With increase in temperature, hydrogen molecules dissociate into hydrogen atoms which are held on the surface by chemisorption.

Assuming ideal behaviour,  $P=\frac{dRT}{M}$

$P=\frac{100}{760}atm,T=257+273=530K$

d = 0.46gm/L

$M=\frac{0.46\times 0.082\times 530}{100}\times 760=151.93\cong 152g/mol$

85gm NH₃ = 5 moles of NH₃

Enthalpy change for 1 mol = 23.4 kJ

Then enthalpy change for 5 mol = 23.4 * 5 = 117 kJ

Mass = d × v = 1.02 × 1.2 = 1.224gm

Moles of acetic acid = 0.0204 moles in 2L

So molality = 0.0102 mol/kg

$\Delta T_f=i\times K_f\times m$

i = 1 + a for acetic acid

0.0198  = (1 + a) × 1.85 × 0.0102

α = 0.04928 = 5%

At anode (oxidation)

2H₂O $\underset{}{\overset{}{\to }}$ O₂ (g) + 4H⁺ + 4e^-

At cathode (Reduction)

2H⁺ + 2e^-$\underset{}{\overset{}{\to }}$ H₂ (g)

No. of gm- equivalents = $\frac{i\times t}{96500}=\frac{0.1\times 2\times 60\times 60}{96500}=0.00746$

$V_{O_2}=\frac{0.00746}{4}\times 22.7=0.0423L$

$V_{H_2}=\frac{0.00746}{2}\times 22.7=0.0846L$ $V_{total}\cong 127mL$

 $\mathcal{l}n\left(\frac{K_2}{K_1}\right)=\frac{Ea}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

$\mathcal{l}n\left(\frac{K_2}{K_1}\right)=\frac{532611}{8.3}\times \left(\frac{10}{310\times 300}\right)$

Where, K₂ is at 310 K and K₁ at 300K

$\mathcal{l}n\left(\frac{K_2}{K_1}\right)=6.9=3\times \mathcal{l}n10$

$\mathcal{l}n\left(\frac{K_2}{K_1}\right)=\mathcal{l}n10^3$

K₂ = K₁ × 10³

K₁ = K₂ × 10^-3

$\therefore$ x = 1

Kindly consider the following figure

 $3Mn{O}_4^{2-}+4H^{+}\underset{}{\overset{}{\to }}2Mn{O}_4^{-}+Mn{O}_2+2H_{2}O$

No. of unpaired electron in Mn⁷⁺ = 0

$\mu =0B.M$

Millie q. of H₂SO₄ used by NH₃ = 12.5 × 1 × 2 = 25

So millimoles of N = 25

Moles of N = 25 × 10^-3

Wt of N = 14 × 25 × 10^-3

% of N =    $\frac{14\times 25\times 10^{-3}}{0.55}\times 100=63.66\cong 64$

No. of compounds containing asymmetric carbon are

Kindly consider the following figure

Moles of Fe₃O₄ =   $\frac{4.64*10^3}{232}=20$

              Moles of CO =   $\frac{2.52*10^3}{28}=90$

              So limiting reagent = Fe₃O₄

              So moles of Fe formed = 60

              Weight of Fe = 60 * 56 = 3360

(A) Cr = [Ar]3d⁵4s¹

              (B) m =   $-\mathcal{l}to+\mathcal{l}$

              (C) According to Aufbau principle, orbital are filled in order of their increasing energies.

              (D) Total nodes = n – 1

Covalent character is $LiC\mathcal{l}>NaC\mathcal{l}>KC\mathcal{l}>CsC\mathcal{l}.$  As the cationic size increases polarization decreases.

Due to common ion effect solubility of AgCl will decreases in KCl, AgCl and AgNO₃ but in deionized water, no common ion effect will takes place so maximum solubility.

Due to equal and similar charge particle will repel each other, hence will never precipitate.

Moles of Fe₃O₄ = $\frac{4.64\times 10^3}{232}=20$

Moles of CO = $\frac{2.52\times 10^3}{28}=90$

So limiting reagent = Fe₃O₄

So moles of Fe formed = 60

Weight of Fe = 60 × 56 = 3360

(A) Cr = [Ar]3d⁵4s¹

(B) m = $-\mathcal{l}to+\mathcal{l}$

(C) According to Aufbau principle, orbital are filled in order of their increasing energies.

(D) Total nodes = n – 1

Covalent character is $LiC\mathcal{l}>NaC\mathcal{l}>KC\mathcal{l}>CsC\mathcal{l}.$ As the cationic size increases polarization decreases.

$\therefore$ Covalent character decreases.

Due to common ion effect solubility of AgCl will decreases in KCl, AgCl and AgNO₃ but in deionized water, no common ion effect will takes place so maximum solubility.

Due to equal and similar charge particle will repel each other, hence will never precipitate

Correct electronic configuration of Pt is ⁷⁸Pt = [Xe]4f¹⁴5d⁹6s¹

It is an exceptional electronic configuration

KCN is used as depressant for ZnS. While Au and Ag, cyanide process (leaching) is used.

In reducing action, H₂O₂ changes to O₂ because it will oxides but option (A) is in acidic medium, hence answer will be (C).

Metals                Li Na K Rb                        Cs

Colour               Crimson red Yellow                 Violet                  Red violet  Blue

Wavelength       670.8  589.2  766.5                   780  455.5

Cs is used in devising photoelectric cells. Boron fibres are used in making bullet – proof vest silicones being surrounded by non-polar alkyl group are water repelling in nature.

Gallium is less toxic and has a very high temperature thermometers

P₄ + 3NaOH + 3H₂O   $\underset{}{\overset{}{\to }}P{H}_3+3Na{H}_{2}P{O}_2$

Oxoacid is H₃PO₂ it is also known as hypophosphorous acid or phosphinic acid.

CaCO₃ + H₂SO_{4 $\underset{}{\overset{}{\to }}$} CaSO₄ + CO₂ + H₂O

C₄H₈ (DOU = 1 so 1 double bond exist since it gives KMnO₄ test)

Buna – N is synthetic rubber which can be stretched and retains its original status on releasing the force.

DNA contains b-D-2- deoxyribose sugar while RNA contains b-D-ribose.

Phenol is weakly acidic compound but it is more acidic than alcohol and water because its conjugate base is more stable. Hence statement -I is correct while statement -II is wrong.

Mass of C₁₅H₃₀ = volume × Density

= 1000 $m\mathcal{l}$  × 0.756 gm/ $m\mathcal{l}$  

C₁₅H₃₀ + 22.5 O₂   $\stackrel{}{\to }$ 15CO₂ + 15H₂O

No. of moles of C₁₅H₃₀ = $\frac{756}{210}$ moles

No. of moles of O₂ required =   $\left(22.5\times \frac{756}{210}\right)moles$

Mass of O₂ required = 22.5 ×   $\frac{756}{210}\times 32=2592gm$

No. of moles of CO₂ liberated = 15 ×   $\left(\frac{756}{210}\right)$ moles

Mass of O₂ liberated =   $15\times \frac{756}{210}\times 44=2376gm$

Degenerate orbitals must have same value of energy

Orbitals with same n and   values are degenerate orbitals.

${\left[PtC{l}_4\right]}^{2-}\to Pt$ has dsp² hybridization

BrF₅ -> Br has sp³d² hybridization

PCl₅ -> P has sp³d hybridization

[Co (NH₃)₆]³⁺ -> Co has d²sp³ hybridization.

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}\times P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}\times P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}\times P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}\times {\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃

Sphalarite           ZnS

Calamine            ZnCO₃

Galena                PbS

Siderite               FeCO₃

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having use in manufacturing of N-based fertizers)

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by lewis base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{\Delta }{\to }}3NaB{F}_4+2B_2{H}_6$                

NF₃ is stable due to high N-F bond energy.

Other halides of nitrogen (NCl₃ NBr₃, Nl₃) are explosive

Enamel has calcium hydroxyl apatite [Ca₁₀ (PO₄)₆ (OH)₂], CaCO₃, CaF₂ and Mg₃ (PO₄)₂

Ca²⁺, P⁵⁺ and F^- are present

Kindly consider the following figure

Melting point of monocarboxylic acids with even number of carbon is more than that with odd number of carbon due to lattice energy.

With increase in molar mass of monocarboxylic acids size of alkyl group (Hydrophobic portion) increases and therefore solubility in water decreases.

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

$Cellulose\underset{}{\overset{H^{+}/H_{2}O}{\to }}\beta -D-Glucose\underset{Water}{\overset{B{r}_2}{\to }}Gluconicacid$

Penicillin is not a broad spectrum artibiotic and is used to treat only certain infections caused by streptococci and staphylococci such as pneumonia.

$N{i}^{2+}+2dmg\to \left[Ni{\left(dmg\right)}_2\right]\downarrow$ Rosy-red precipitate

Unit cell = Hexagonal close packing (hcp)

No. of X in one unit cell = 6

No. of tetrahedral voids in one unit cell = 12

No. of Y in one unit cell = $\frac{2}{3}*12=8$ $\frac{}{}$

Formula => X₆ Y₈ => X₃ Y₄

% of X in unit cell $=\frac{3}{7}*100\mathrm{\%}=43\mathrm{\%}$ $\frac{}{}$

2O₃(g) -> 3O₂

t = 0      a moles               0

-0.5a mole     +0.75a mole

At Eq.    0.5a mole         0.75a mole

Total moles at eq = 0.5a + 0.75 a = 1.25a

$P_{O_3}=x_{O_3}\times P_T=\frac{0.5a}{1.25a}\times 1=\frac{2}{5}atm$

$P_{O_2}=x_{O_2}\times p_T=\frac{0.75a}{1.25a}\times 1=\frac{3}{5}atm$

$K_p=\frac{{\left(P_{O_2}\right)}_{eq}^3}{{\left(P_{O_3}\right)}_{eq}^3}=\frac{{\left(\frac{3}{5}\right)}^3}{{\left(2/5\right)}^2}=1.35atm$

$\Delta G^{°}=-RT\mathcal{l}nKP$

$\begin{array}{l}=-8.3\times 300\mathcal{l}n1.35\\ =-747J/mole\end{array}$

For isotonic solution

${?}_{injector}={?}_{Blood}$

$?C*0.082*300=7.47$

$?C=0.3\text{?}\text{?}mole/\mathcal{l}$

$=0.3*180gm/\mathcal{l}=54gm/\mathcal{l}$

Anode, H₂ (g) -> 2H⁺ + 2e^-

Cathode, Cu²⁺ + 2e -> Cu (s)

Net cell reaction H₂ + Cu²⁺ -> 2H⁺ Cu (s)

$E_{cell}^0=E_{C{u}^{2+}/Cu}^0-E_{H^{+}/H_2}^0$

= 0.34 V – 0

= 0.34 V

$E_{cell}=E_{cell}^0-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{\left[C{u}^{2+}\right]}$

$\begin{array}{l}\Rightarrow 0.576=0.34-\frac{0.06}{2}log\frac{{\left[H^{+}\right]}^2}{0.01}\\ \Rightarrow -log\left[H^{+}\right]=4.93\\ \Rightarrow pHofsolution=4.93~5\end{array}$

$k=A.e^{-\left(\frac{Ea}{RT}\right)}$

$\mathcal{l}nk=\mathcal{l}nA-\frac{Ea}{RT}$

$\Rightarrow \mathcal{l}nk=\mathcal{l}nA+\left(-\frac{Ea}{1000R}\right).\frac{1000}{T}$

Slope = $\frac{-Ea}{1000R}=-18.5$  

=> Ea = 18.5 × 1000 × 8.31 = 153.735 × 10³ J = 154 KJ

Kindly consider the following figure

Number of Co-Co Bond = X = 1

Number of terminal ligand = Y = 6

Meq of NH₃ = Meq of used H₂SO₄ = Meq of NaOH = 0.25 × 30 = 7.5

Millmoles of N = millimoles of NH₃ = 7.5 (As n factor = 1)

Mass of nitrogen = 7.5 × 14 × 10^-3 = 0.105 gm

% of Nitrogen = $\frac{0.105}{0.166}\times 100\mathrm{\%}=63.25\mathrm{\%}~63\mathrm{\%}$

Kindly consider the following figure

Number of sp² hybridized carbon in A = 2