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

1. Ans:

(a). According to Ellingham diagram, the reaction of CO2 is more feasible at temperatures lower than 710 K and thus it is a better reducing agent below 710 K.

While the reaction of CO is more feasible at temperatures higher than 710 K and thus it is a better reducing agent at above 710 K.

(b). According to the Ellingham diagram, the more negative the Gibbs free energy of a particular reaction the more feasible it is to carry out. Since the oxides are easier to reduce, sulfide ores are converted into oxides before reduction.

(c). To extract copper, its sulfide ores are supposed to be heated in a reverberatory furnace. The sulfide ore may contain iron as an impurity. So, it is mixed with silica before heating. The oxide of iron slags off as iron silicate. In other words, the iron impurity is removed easily in the form of iron silicate.

(d). Carbon and hydrogen are good reducing agents. They are used preferably for reduction of oxides. They also form escapable gasses as by-products. However, at higher temperatures they form carbides and hydrides. Also, it is not convenient to maintain such high temperatures. Thus, electrolysis is used for this purpose.

(e). Ti is heated in an evacuated vessel along with iodine. Ti forms volatile compounds with iodine. The volatile compound formed decomposes easily. This makes the recovery of the metal convenient. Ti thus fulfills the requirements needed for carrying out vapor phase refining. Hence, this method is used for its purification.

4. Iron oxides are mixed with limestone and coke in order to decompose carbonates and oxidize sulfides. This mixture is then fed into the blast furnace. Many reactions take place in different temperature ranges of the blast furnace. By oxidizing impurities from cast iron in a reverberatory furnace which is lined by hematite, wrought iron or malleable iron is obtained. This is the purest form of commercial iron. This can be shown with the following reaction:

Fe2O₃ + 3C → 2Fe + 3CO.

Sulfur, silicone and phosphorus are oxidized and passed into the slag and removed as impurities when limestone is added as a flux to it.

6. Both processes have two basic requirements:

(i) The metal should react with an available reagent to form a volatile compound.

(ii) The volatile compound should be easily decomposable, allowing for easy metal recovery.

12. Sulfide ore of copper contains some types of impurities. Sulfide ores are not reduced easily, but oxide ores are easily reduced. It is necessary to make sulfide ores free from volatile impurities. This can be performed by a roasting method. In this moisture escape and impurities like sulfur, phosphorus, arsenic are oxidized to their volatile oxides. It is performed in a reverberatory furnace. After this process the mass becomes porous.

15. In metallurgical processes, Flux works as various types of agents like chemical agents, cleaning/purifying agents.

Fluxes have many important properties such as corrosivity, cleanability, conductivity, volatility, etc.

Flux, when combined with gangue together, forms slag i.e. Flux + Gangue = Slag

As compared to gangue, slag separates much easily from the ore, in this way the removal of gangue also becomes much easier.

Flux is useful for making the molten mass more conducting which helps to remove impurities from metals through electrolytic metallurgical processes.

17. The chemical reactions taking place in Blast furnace in the temperature range 500-800 K are:

3Fe₂O₃ + CO→ 2Fe₃O₄ + CO₂

Fe₃O₄ + 4CO→ 3Fe + 4CO₂

Fe₂O₃ + CO→ 2FeO + CO₂

53.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: Froth Flotation method is used to separate the hydrophobic materials from hydrophilic materials. In this method a mixture of palm oil, water and detergent is taken in a tank along with powdered sulfide ore. Compressed air is then passed through the pipe of the rotating agitator to create froth. The sulfide ore is then wetted by the palm oil mixture and it rises with the froth and the impurities or gauge settles at the bottom of the tank. The froth containing the sulfide is then cleaned and dried. Cresols or Aniline are used to stabilize the froth and in case of sulfides palm oil is used to collect non wet particles of the sulfide.

54.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: The zone refining method is very useful for producing high-purity semiconductors and other metals, such as germanium.

2. Extraction of chlorine from brine is an oxidation method. The reactions involved are:

2Cl⁻ + 2H₂O → 2OH⁻ + H₂ + Cl₂

For this reaction, the value of ΔG°=+422 kJ, which is positive. Using the formula ΔG°=−nE°F, we get a negative value of E° =−2.2 V.

Since,  E°=−2.2 V, it would naturally require an external voltage greater than 2.2 V for the reaction to occur.

3. From Ellingham diagram, we know that ΔG^° _{(C, CO)}  < ΔG^° _{(Fe, Fe)}, the following reactions are:

C+ $\frac{1}{2}$ O₂→CO

2Fe+O₂→2FeO

Therefore, FeO can be reduced to Fe by coke. 

5. Low-grade copper ores contain a small percentage of copper that is 0.27% copper. By hydrometallurgy, low-grade copper is leached by treating with acid when copper metal goes into solution as copper ions. The solution containing copper ions is treated with scrap iron or hydrogen gas. Since iron or hydrogen is more reactive than copper so they reduced copper ions from solution to copper metal.

7. The reduction is the removal of oxygen. Hydrogen and carbon both form oxides that are easy to remove both water and carbon dioxide which are produced as a gas at hot reactions. Although carbon and hydrogen are better reducing agents, they are not used to reduce metallic oxides at high temperatures because at high temperatures carbon and hydrogen react with metals to form carbide and hydrides respectively.

8. Separation of two sulfide ores can be accomplished by adjusting the proportion of oil to water or by using depressants. In the case of an ore containing ZnS and PbS, for example, the depressant NaCN is used. It forms a complex with ZnS and prevents it from coming into contact with froth, but PbS remains in contact with froth.

9. A metal is called Pure when it contains the highest concentration of itself and the least concentration of impurities. While refining of metal, these impurities are removed by various processes aided by the use of other compounds.

Hematite supplies oxygen and oxidizes carbon, silicon, manganese, and phosphorus present in the cast iron to carbon monoxide, manganese oxide, phosphorus pentoxide respectively.

Fe₂O₃ + 3C  2Fe + 3CO

3Si + 2Fe₂O₃ → 4Fe + 3SiO₂

3S + 2Fe₂O₃→3SO₂ + 4Fe

Wrought iron is called the purest form of iron as it contains the highest percentage of Iron and the least impurities. It is prepared from cast iron by oxidizing impurities in a reverberatory furnace lined with Hematite.

10. Column chromatography is a technique used to separate mixtures based on differences in their affinity towards the stationary phase. If one molecule in a solute has more affinity towards the stationary phase, it would form bonds with the stationary phase and hence remain in the column, Whereas, if the other one has less affinity, it would not form bonds and hence will travel down and come out of the column faster.

As compound A is eluted in preference to compound B, compound B is more readily adsorbed on the column of Al₂O₃.

10. Sulfide ores are not reduced easily, but oxide ores are easily reduced. Sulfide ore of copper contains some types of impurities. This can cause by-product formation and degrade our final product. Iron oxide is present as a key impurity of concern in sulfide ore of copper. During the roasting process the temperature in the furnace is near about 1200°C. During such high temperatures iron oxide which is gangue (impurity) in ore forms slag. Slag is insoluble in molten metal and being lighter floats over the surface of molten metal. Here silica is used as flux.

FeO + SiO₂ →FeSiO₃ 

13. Vapor phase refining or Van Arkel's method is used for refining Zr and Ti.

Vapor refining or vapor phase process: The metal is collected in the form of volatile compound that decomposes to pure metal.

Van Arkel method: This method is useful for removing all Oxygen and nitrogen present in the form of impurity in metals like Zr and Ti. The crude metal is heated in an empty vessel with a small amount of iodine. The metal iodine being more covalent volatilises. Impure Zr reacts with iodine to form zirconium tera iodide. 

The following reaction occurs Zr + 2I₂? ZrI₄ The metal iodide is decomposed on a tungsten filament electrically heated to about 1800 K. The pure metal is deposited on the filament and to maintain temperature the current is steadily raised, as the depositing continues.

14. The considerations during the extraction of metals by electrochemical method are :

a) Reactivity of metals: If the metals are reactive and are likely to react with water then the metals should be isolated by the electrolysis of their purified molten ore rather than their aqueous solution.

b) Suitability of electrodes:The electrode is carefully chosen so that they should not react with the product of electrolysis. If they react then the electrodes must be made of a material which is quite cheap so their replacement should not increase the cost of the process.

16. Semiconducting metal is produced by the zone refining method.

Zone refining is centered on the principle that the impurities which are included in the metal ores are more soluble in the molten state as compared to the solid-state of metals.

Zone refining is mostly used for obtaining semiconductor and other metals which are highly pure, e.g., germanium, silicon, etc.

18. The two import requirements for performing the Vapor Phase Refining Method:

a) The metal which is to be refined should form a volatile compound with an available reagent.

b) The volatile compound formed must be easily decomposable so that it is easy to recover the desired metal.

19. The metal gold is a nonreactive metal. It is found as a native metal in its free state.

So, it is not compulsory to separate it by a chemical process. Thus, gold is leached to get it in its pure form. In the metallurgy of gold (Au), the respective metal is leached with a dilute mixture of NaCN or KCN in the presence of air where Gold (Au) is oxidized by oxygen of the air to Au+cation which then merges with ions to form a soluble compound.

4Au (s)+8CN⁻ (aq)+O₂⁻ (g)+2H₂O⁻ (aq)→4 [Au (CN)₂]⁻ (aq)+4OH⁻ (aq)

Gold is then obtained from this soluble compound x by replacement method by using a much more electropositive Zinc metal.

In this method of Gold extraction, the metal Zinc (Zn) plays the role of a reducing agent and it reduces Au+to Au.

Zn itself gets oxidized to Zn+ions which combine with CN−ions to form another soluble complex. 2Na [Au (CN)₂] (aq)+Zn→2Au+Na₂ [Zn (CN)₄].

20. Option (i)

Explanation: 2H₂O + Cl⁻→2OH⁻ + H₂ + Cl₂

The ΔG^?  is +422 kJ for this reaction. We get ΔE^? = −2.2 V, when we convert it to ΔE^?   (using G^? = E^? F). It will, of course, require an external e.m.f. greater than 2.2 V. However, electrolysis necessitates an excess potential in order to overcome some other impeding reactions. Thus,  Cl₂ is obtained through electrolysis, which produces H₂ and aqueous NaOH as byproducts. In addition, molten NaCl is electrolyzed. However, in that case, Na metal is formed rather than NaOH.

21. Option (iii)

Copper ore is mixed with silica before heating in a reverberatory furnace. When heated in a reverberatory furnace, the iron oxide slags of iron silicate and copper are produced. This product is produced in the form of copper matte in the furnace. Presence of Cu₂S and FeS is observed in copper matte. In these compounds, we observe the sulfides of Copper (I) Cu₂⁺ and Iron (II) Fe₂⁺.

22. Option (iv)

Copper (I) oxide is reduced by copper (I) sulfide in this reaction. Because copper is reduced by itself in this process, it is referred to as auto reduction.

23. Option (i)

Explanation: Aluminum is the third most abundant metal in the earth's crust (8.3 percent approximately by weight). It is found in a variety of igneous minerals, including mica and clays. The second most abundant metal in the earth's crust is iron.

24. Option (ii)

Purification of metal crystals by making a thin region of crystal undergo melting is known as Zone refining. The molten crystal is then moved up along the crystal to get pure form of it. This process is used to get pure form of Silicon and Germanium. Basic principle of the zone refining process is that the impurities are more soluble in molten metal than solid metal to get pure form of metal.

25. Option (iii)

Reaction takes place as Cu₂O + $\frac{1}{2}$ Cu₂S?3Cu +12SO₂ with a product as bristle copper. This type of reaction is called an auto-reduction reaction as copper is reducing itself with help of other copper compounds.

26. Correct option (i)

The reaction at the anode with a lower E $°$  value is preferred, but oxygen cannot be obtained in this process due to overvoltage.

27. Option (ii)

Reaction involved in the metallurgy of aluminum is 2Al₂O₃ + 3C→ 4Al + 3CO₂.

The reaction at cathode is Al³⁺ + 3e⁻→ Al

The reaction at anode is

C+12O₂ →CO + 2e and C+O₂→CO₂ + 4e⁻

Hence, from the reaction graphite anode is oxidized to carbon monoxide and carbon dioxide.

28. Option (i)

The electrolytic method can be used to purify zinc and copper. The impure metal is used as an anode in this method. As the cathode, a pure strip of the same metal is used. They are immersed in an appropriate electrolytic bath containing a soluble salt of the same metal.

The more basic metals remain in the solution, while the less basic metals are transferred to the anode mud.

29. Option (i) 

The cyanide process involves 3 steps:

First step - The finely grounded ore of gold and silver are made to come in contact with the solution containing the cyanide,

Second step - it involves separation of gold and silver from the cyanide solution  

Third step - it involves the recovery of gold and silver in their pure forms precipitating the remaining solution with zinc dust.

Thus, the metal is recovered by displacing Zn with the metal (Au or Ag) from metal ions.

30. Option (iv)

Above point A, ΔG (C, CO)<ΔG (Fe, FeO). Thus, carbon reduces FeO to iron i.e. Fe and in order carbon is oxidized to carbon monoxide i.e. CO. Both these lines intersect at Point A. As we can see in the figure above point A only carbon could reduce FeO in Fe and CO is being produced.

31. Option (i)

Below point A, only the value of ΔG (CO, CO₂) is less than the value of ΔG (Fe, FeO) at the corresponding temperatures. Thus, only carbon monoxide will be able to reduce FeO to Fe and will get itself oxidized into CO₂.

32. Option (i)

According to the above graph, at point D the equivalent value of ΔG for the reduction of FeO is approximately 330 units for the particular temperature. Also, for CO the equivalent value of Δ?  at the particular temperature for point D  is around −330 units. Thus, if we calculate the overall value of ΔG for the reduction of FeO with carbon monoxide then it is almost zero at point D.

33. Option (ii) and (iv)

From Fig.6.1, at point B, all three lines, i.e. for the lines for all three given reactions, are below that of the line of reduction of FeO. This means, that the value of ΔG for the equivalent temperature at point B for reduction of FeO is greater than (positive) that of values of ΔG for all three reactions (negative). Thus,  FeO will get reduced by all three reactions at point B. Same is the case with point E as that of point B, the value of ΔG for reduction of FeO at E is greater than the value of ΔG for all three reactions at the equivalent temperature. Thus,  FeO will get reduced by all three reactions at point E.

34. Option (i) and (iv)

Pig iron is the iron obtained from the blast that has many impurities (Mn, P, Si, etc.) and approximately 4% carbon content. When it is re-melted or cast with the help of scrap iron, then it has less carbon content i.e. 3% as compared to pig iron.

The Zr andTi are purified by the help of Arkel Method that contain the following reaction: 

Zr (s) + 2I₂   ZrI₄ ( g)  Zr Pure  (s) + 2I₂

Ti + 2I₂  TiI₄  Ti + 2I₂

Here we obtain Zr and Ti as a pure form of solid when purified using the Arkel method.

35. Option (i) and (ii)

In the extraction of alumina by Hall-Heroult's process, alumina is mixed with cryolite with the mixture of fluorspar \& fluoride that increases the conductivity of electrical and lowers the melting point of Al₂O₃ .

36. Option (i) and (iii)

A suspension of powdered ore is made with water during the froth floatation process. Collectors and froth stabilizers are added to it. Collectors (e.g., pine oils, fatty acids, xanthates, etc.) improve the non wettability of the mineral particles, while froth stabilizers (e.g., cresols, aniline) keep the froth stable 

37. Option (ii) and (iii)

It is possible to separate two sulfide ores using the froth floatation process by adjusting the proportion of oil to water or by using 'depressants.' In the case of an ore containing ZnS and PbS, for example, the depressant used is NaCN.

38. Option (iii) and (iv)

Bauxite consists of many impurities like hematite, goethite,  Fe₂O₃, the sand SiO₂, etc.'

39. Option (ii) and (iii)

The froth floatation method is most commonly used for sulfide ore. Sulfide ores galena (PbS) and copper pyrites  (CuFeS₂) are found here.

40. Option (i) and (iii)

Calcination is the process of heating when the volatile matter escapes, leaving the metal oxide behind. It is usually done in the absence of air.

41. Option (i) and (iii)

Explanation: Haematite is an ore of iron that can be calcined and reduced by carbon. In the metallurgical process, calamine ore is calcined ore that can be reduced by carbon. Hence, option (i) and (ii) are correct.

42. Option (i) and (iv)

(i) Carbon monoxide is the primary reducing agent in the furnace.

(ii) This is an endothermic reaction in which heat is absorbed from the furnace. As a result, it is critical not to add too much limestone, as this will cool the furnace. Calcium oxide is a basic oxide that reacts with acidic oxides in the rock, such as silicon dioxide. Calcium silicate is formed when calcium oxide reacts with silicon dioxide.

43. Option (i) and (ii)

In the vapor phase refining method, the metal is converted into its volatile compound and then is collected elsewhere. This involves two techniques:

1. Mond Process for refining Nickel: in this process, a volatile complex, nickel tetracarbonyl is formed when nickel is heated with a stream of carbon monoxide.

2. Van Arkel Method for refining Zirconium or Titanium: this method is basically used for removal of oxygen and nitrogen present as impurities in the Zr or Ti metal. These are heated in an evacuated metal with iodine. As iodine is more covalent than these metals, it volatilizes.

44. Option (i) and (ii) 

Explanation: Depressants are materials that are added for the separation of ores that prevent certain types of particles from coming to froth and forming bubbles. For example, an ore containing ZnS and PbS, NaCN is used as a depressant.

When sulfur ores are blown in hot air along with silica, the solidified metal which is obtained has a blistered appearance due to SO₂ evolution.

45. Option (ii) and (iii)

Explanation: Using oxidation method for extraction of chlorine from brine. The reactions involved are:

2Cl⁻ + 2H₂O → 2OH⁻ + H₂ + Cl₂

For this reaction, the value of ΔG°=+422 kJ, which is positive. Using the formula ΔG°=−nE°F, we get a negative value of E° =−2.2 V.

46. Option (ii) A (2) B (4) C (5) D (3)

Pendulum is always made of nickel steel.

Hence, option (A) from column I is matched with option (2) from column II. Malachite is the ore of copper.

Hence, option (B) from column I is matched with option (4) from column II. Calamine is the ore of zinc.

Hence, option (C) from column I is matched with option (5) from column II. Cryolite is an ore of aluminum.

Hence, option (D) from column I is matched with option (3) from column II.

47. Option (ii) A (4) B (3) C (1) D (2)

Coloured bands are found in chromatography.

Hence, option (A) from column I is matched with option (4) from column II. Impure metals are converted to volatile complexes in Mond's process.

Hence, option (B) from column I is matched with option (3) from column II. Purification of Ge and silicon is done using zone refining.

Hence, option (C) from column I is matched with option (1) from column II. Purification of mercury is done using fractional distillation.

Hence, option (D) from column I is matched with option (2) from column II.

48. Correct code (i) A (4) B (2) C (3) D (1)

The cyanide process is used in the extraction of Au.

Hence, option (A) from column I is matched with option (4) from column II. Froth floatation process is used in the dressing of ZnS.

Hence, option (B) from column I is matched with option (2) from column II. Electrolytic reduction is used in the extraction of AI.

Hence, option (C) from column I is matched with option (3) from column II. Zone refining is used to get ultrapure Ge.

Hence, option (D) from column I is matched with option (1) from column II.

49. Option (i) A (3) B (4) C (2) D (1)

Sapphire is a gemstone containing Co.

Hence, option (A) from column I is matched with option (3) from column II. The Sphalerite single is ZnS.

Hence, option (B) from column I is matched with option (4) from column II. NaCN is also used as a depressant.

Hence, option (C) from column I is matched with option (2) from column II. Al2O3 is also called corundum.

Hence, option (D) from column I is matched with option (1) from column II.

50. Option (i) A (2) B (3) C (4) D (1)

Explanation: A solidified copper has a blistered appearance due to the evolution of SO2 Hence it is called blistered copper.

Hence, option (A) from column I is matched with option (2) from column II. Iron is extracted from a blast furnace.

Hence, option (B) from column I is matched with option (3) from column II. The iron ore is heated in the reverberatory furnace after mixing with silica. In the furnace, iron oxide slags of iron and copper are produced in the form of copper matte.

Hence, option (C) from column I is matched with option (4) from column II. The hall-Heroult process is used for the extraction of aluminum from its ore.

Hence, option (D) from column I is matched with option (1) from column II.

51.  (i) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: In the Mond process Nickel is reacted with carbon monoxide reversibly to give Nickel carbonyl,  Ni (CO)₄. Nickel carbonyl is a volatile compound. It decomposes to nickel and carbon monoxide at 460 K.

52. (i) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation: Van Arkel method is generally used to obtain pure forms of Zirconium (Zr) and Titanium (Ti).

The metal iodide is heated at 1800 K and is decomposed on a tungsten filament. The pure metal is dropped on the tungsten filament. This proves that ZrI4 is volatile and decomposes at 1800 K.

55.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: Hydrometallurgy is used to extract copper from low-grade ore. Hydrometallurgy entails dissolving the ore in a suitable reagent and then precipitating it. In this method, more electropositive metal is used, allowing pure metal to be displaced.

 $Molality=\frac{molesofsolute\times 1000}{wtofsolvent\left(gm\right)}$

$Molality=\frac{35\times 1000\times 1.46}{36.5\times 100}$ = 14.0 M

$Bondorder=\frac{e^{-}ofBondingMO-e^{-}ofABMO}{2}$

Hence Bond order for $O_2^{+}=\frac{10-5}{2}=2.5$

$O_2=\frac{10-6}{2}=2.0$

$O_2^{-}=\frac{10-7}{2}=1.5$

$O_2^{--}=\frac{10-8}{2}=1.0$

According to Ellingham diagram, the metal oxide with lower $\Delta G^o$ is more stable than metal oxide of higher $\Delta G^o$ at same temperature.

Due to H- bond in water, it has high melting point and melting point of other hydrides of the group are depending upon the molecular weight.

This is the formation of free radical mechanism, hence free radical will be formed

Buna-S is the copolymer of butadiene and styrene, Nylon-6, 6 is the condensation polymer of adipic acid and diamine and having fibrous nature.

Brown ring is formed due to nitrosoferrous sulphate formation.

Wt of liquid = 135 – 40 = 95 gm

Volume of liquid = $\frac{wt.}{density}=\frac{95}{0.95}=100ml$

Hence volume of vessel = 100 ml = 0.1 lit from ideal gas equation,

$M=\frac{dRT}{P}=\frac{0.5}{0.1}\times \frac{0.0821\times 250}{0.82}$

$\therefore M=125g/mol$

Radius of Bohr’s orbit $R_n=0.529\stackrel{o}{A}\times \frac{n^2}{z}$

Radius of Bohr’s orbit for hydrogen,  $R_n=0.529\stackrel{o}{A}\times n^2$

For third orbit (R₃) = $0.529\stackrel{o}{A}\times 9$ = r₃ and

Fourth orbit (R₄) = $0.59\stackrel{o}{A}\times 16=r_4$

$\therefore r_4=\frac{16r_3}{9}$

 $\frac{dE}{dT}$ is the temperature Co-efficient of cell. The cell having less variation of EMF, with respect to temperature have high efficiency.

During the electrolysis of dilute H₂SO₄

$2H_{2}S{O}_4\stackrel{\left(1\right)electrolysis}{\to }2HS{O}_4^{-}+2H^{+}$

$H_2{S}_2{O}_8+2H_{2}O\stackrel{hydrolysis}{\to }2H_{2}S{O}_4+\underset{\left(A\right)}{H_2{O}_2}$

$2H^{+}+2e\to H_2\uparrow$

In the solid form of $H_2{O}_2$ dihedral angle is equal to 90.2°.

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

$P_4+3NaOH+3H_{2}O\to P{H}_3+3Na{H}_{2}P{O}_2$

$RedP+alkali\to \underset{\left(NoP-Hbond\right)}{H_4{P}_2{O}_6}$

HOCl produce in the stratospheric cloud, by the hydrolysis reaction of ClONO₂.

$ClON{O}_{2\left(g\right)}+H_2{O}_{\left(g\right)}\to HOC{l}_{\left(g\right)}+HN{O}_{3\left(vap\right)}$

Regarding Lassaigne's Test, both statement are correct.

Aromaticity drives the highest enolic percentage of given structure:

Enamines are inter conversible and have low stability with respect to imine. Among all C is most stable due to steric factor.

Histamine is the chemical which stimulates the secretion of pepsin in stomach.

 $C{H}_{3}O{H}_{\left(\mathcal{l}\right)}+\frac{3}{2}{O}_{2\left(g\right)}\to C{O}_{2\left(g\right)}+2H_{2}O\left(\mathcal{l}\right)$

$\Delta n=-0.5$

$\Delta H=\Delta E+\Delta nRT=-726-0.5\times \frac{8.3}{1000}\times 300=-726-1.24=-727.24\approx -727kJ/mol$

Hence, x = 727 (the nearest integer)

0.5 % KCl solution has molality (m) = $\frac{0.5\times 1000}{74.5\times 99.5}$

$KC{l}_{\left(aq\right)}?K_{\left(ag\right)}^{+}+C{l}_{\left(aq\right)}^{-}$

1 - α            α             α

And I = $\left(1-\cancel{\alpha }+\cancel{\alpha }+\alpha \right)=1+\alpha$

$i=\frac{\Delta Tf}{kf\times m}=\frac{0.24\times 74.5\times 99.5}{1.8\times 0.5\times 1000}=1+\alpha$

1.976 = 1 + α

$\therefore \alpha =0.976$

% = 97.6%

the nearest 98.

Here, total meq of acetic acid = 50 × 0.1 = 5

And total meq of NaOH = 25 × 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$pH=pKa+lo{g}_{10}\frac{\left[S\right]}{\left[A\right]}$

$pH=4.76+lo{g}_{10}\frac{2.5}{2.5}=4.76=476\times 10^{-2}$

Process is based upon simultaneous disintegration hence,

$\frac{0.693}{100}\times t=2.303lo{g}_{10}\frac{A_0}{A_t}$ ………….(i)

and $\frac{0.693}{50}\times t=2.303lo{g}_{10}\frac{B_0}{B_t}$ ………….(ii)

from equation (i) and (ii)

$0.693t\left[\frac{1}{50}-\frac{1}{100}\right]=\left[log\frac{B_0}{B_t}-log\frac{A_0}{A_t}\right]\times 2.303$

Here; A0 = B0 and $A_t=4\times B_t$

Therefore $0.693t\left[\frac{1}{100}\right]=2.303\left[log\left(\frac{B_0}{B_t}\times \frac{A_t}{A_0}\right)\right]$

$\therefore t=\frac{2.303\times 0.3010\times 2\times 100}{693}=200s$

Volume of H₂ adsorbed = $\frac{nRT}{P}=\frac{2\times 0.083\times 300}{2\times 1}=24.9lit=24900ml$

Therefore volume of gas adsorbed per gram of the adsorbent = $\frac{24900}{2.5}=9960$

Most basic oxide V₂O3

Here V has +3 O.S. Hence V⁺³ $\left[Ar\right]3d^2$

two unpaired e^- in d- subshell

$\mu =\sqrt[]{2\left(2+2\right)}=\sqrt[]{8}=2.84BM\approx 3$

For precipitation of two moles of AgCl

Two Cl will produce as a free anion

CoCl₃.4NH₃ complex will $\left[CO{\left(N{H}_3\right)}_{4}C{l}_2\right]$ Cl (will not give 2Cl)

$PtC{l}_4.2HCl\to$ complex will be H2 [PtCl6] will not any Cl

$NiC{l}_2.6H_{2}O\to \left[Ni{\left(H_{2}O\right)}_6\right]C{l}_2$ will produce two Cl ion.

${\left[Ni{\left(H_{2}O\right)}_6\right]}^{++}+2C{l}^{-}\underset{AgN{O}_3}{\overset{}{\to }}2AgCl\left(s\right)$ precipitate formation

$N{i}^{2+}\to \left[Ar\right]3d^{8}4s^0$

$\mu =\sqrt[]{2\left(2+2\right)}=\sqrt[]{8}=2.84BM=3$

Complete combustion of compound produces 0.2 gm CO₂

Hence wt of carbon in 0.2 gm CO₂

$=\left(\frac{12}{44}\times 0.2\right)gm$

Therefore % of carbon in compound

$=\frac{wtofcarbon\times 100}{wtofcompound}=\frac{12\times 0.2\times 100}{44\times 0.3}=\frac{2400}{44\times 3}=18.18\approx 18\mathrm{\%}$

Intermolecular H- bonding and intra-molecular H- bonding producing compound may be the phenol derivatives.

Consider the following image

$N{O}_3^{-}+4H^{+}+3e^{-}\to NO+2H_{2}O;E_{N{O}_3^{-}}^0=0.96V$

$Let\left[HN{O}_3\right]=yM$

For same thermodynamic tendency of reduction

$E_{N{O}_3^{-}/NO}=E_{N{O}_3^{-}/N{O}_2}$

$0.96-\frac{0.059}{3}log\frac{p_{NO}}{y^5}=0.79-0.059log\frac{p_{N{O}_2}}{y^3}$

Assume $\left(p_{NO}=p_{N{O}_2}=1bar\right)$

$0.17=-\frac{0.0591}{1}log\frac{p_{N{O}_2}}{y^3}+\frac{0.059}{3}log\frac{p_{NO}}{y^5}$

$=\frac{0.059}{3}\left[log\frac{p_{NO}}{y^5}-log\frac{{p^3}_{N{O}_2}}{y^9}\right]=\frac{0.059}{3}\times log\left[\frac{p_{NO}}{y^5}\times \frac{y^9}{{p^3}_{N{O}_2}}\right]=\frac{0.059}{3}log{y}^4$

$log{y}^4=\frac{0.17\times 3}{0.059}\Rightarrow y=10^{2.16}$

Comparing x = 2.16 and 2x = 4.32

Nearest integer is 4.

$log\left(\frac{k_2}{k_1}\right)=\frac{Ea}{2.303\times 8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

log (5) = $\frac{Ea}{2.303\times 8.314}\left[\frac{1}{300}-\frac{1}{325}\right]$

$\therefore Ea=\frac{0.7\times 2.303\times 8.314\times 300\times 325}{25}=52271.7$

Ea in kJ/mole = $\frac{52271.7}{1000}=52.2kJ/mol$

the nearest integer is 52.

Density = $\frac{M\times z}{N_A{a}^3}$

$d=\frac{63.5\times 4}{6.02\times 10^{23}\times {\left(3.596\times 10^{-10}\right)}^3\times 1000}=9076kg/m^3$

If $\Delta E=0$

Q = W

W = -P_ext $\left(\Delta V\right)=-4.3nRT\left[\frac{1}{P_2}-\frac{1}{P_1}\right]$

= $-4.3\times 5\times 8.314\times 293\left[\frac{1}{1.3}-\frac{1}{2.1}\right]$

= 15347.70 K = 15.3 kJ

Q = 15 kJ

$AB?A^{+}+B^{-}$

1-x x x   $\therefore i=1+x$

$\begin{array}{l}\Delta T_b=i\times k_b\times m\\ 2.5=\left(1+x\right)\times 0.52m\end{array}$

$\therefore molality=\frac{2.5}{1.75\times 0.52}=2.74$

the nearest integer is 3.

M_eq of NaOH = M_eq H₂C₂O₄

4.44 × N = 1.25 × 2 × 10

$\therefore N=\frac{1.25\times 2\times 10}{4.44}=5.63$

$\therefore N=M=5.63$

Please find the below image

$E=\frac{hc}{\lambda }=\left\{\frac{6.63\times 10^{-34}\times 3\times 10^8}{663\times 10^{-9}}\right\}$

= 0.03 × 10^-17 = 3.0 × 10^-19 J/atom

= 18.06 × 10¹ kJ/mole = 180.6 181 KJ/mole

Cu⁺ = 3d⁹ 4s⁰

n = 1

$\mu =\sqrt[]{1\times \left(1+2\right)}=\sqrt[]{3}=1.73BM$

the nearest integer is 2.