Chemistry NCERT Exemplar Solutions Class 11th Chapter Ten: Overview, Questions, Preparation

This is a assertion and reason type question as classified in NCERT Exemplar

(A) Lithium beimg small in size, polarises large carbonate ion. Polarisation is the distortion of electron cloud of the anion by the cation. Thus, the carbonate of lithium decomposes easily on heating to form lithium oxide and CO₂. The reaction is shown below.

Li₂CO₃ → Li₂O+CO₂

This is a short answer type question as classified in NCERT Exemplar

In the solid state BeCl₂ has a polymeric chain like structure. Be atom is surrounded by four Cl atoms among which two of them are bonded through covalent bond and other two are through co-ordinate bond. Structure of BeCl₂ is given by structure A.

In gaseous state beryllium chloride exists as dimer (Be₂Cl₄) which dissociates to the monomer at about 1200 K temperature which is linear in structure.

Structure of BeCl₂ in gaseous state is given by the structure B.

This is a long answer type question as classified in NCERT Exemplar

(i) Alkaline earth metal compounds are less ionic than alkali metals because of small size and more effective nuclear charge.

(ii) Oxides of alkali metals are more basic than alkaline earth metals. This are water soluble and highly exothermic. The hydroxides of alkaline earth metals are less basic than alkali metals.

(iii) Alkaline earth metals give oxosalts. The reactivity of alkali metals is faster than the reactivity of alkaline earth metal. The reactivity of alkaline earth metal is less due to small size and more effective nuclear charge.

(iv) The oxo salts of alkali metals are less soluble than the oxo salts of alkaline earth metal because of small size of cation and high hydration enthalpy.

(v) The thermal stability of the oxo salts of alkali metals are more than the alkaline earth metals. The sodium carbonate is stable towards heat.

This is a short answer type question as classified in NCERT Exemplar

The lewis structure of O₂^– ion is,

Oxygen atom having no charge has 6 electrons, so its oxidation number is zero. Oxygen atoms containing −1 charge have 7 electrons, so its oxidation number is −1. The average oxidation state of oxygen in this ion is, =1/2

This is a matching type question as classified in NCERT Exemplar

(i) → (f); (ii) → (d); (iii) → (b) : (iv) → (c) ; (v) → (e); (vi) → (a)

All alkali metal and alkaline earth metals except beryllium and magnesium gives characteristic color when introduced into flame. Due to released energy being absorbed in the visible region, alkali metals and alkaline earth metals gives characteristic color.

(i) Caesium imparts blue color when introduced into flame.

(ii) Sodium imparts yellow color when introduced into flame.

(iii) Potassium imparts violet color when introduced into flame.

(iv) Calcium imparts brick red color when introduced into flame.

(v) Strontium imparts crimson red color when introduced into flame.

(vi) Barium imparts apple green color when introduced into flame.

(iv) Calcium imparts brick red color when introduced into flame.

(v) Strontium imparts crimson red color when introduced into flame.

(vi) Barium imparts apple green color when introduced into flame.

This is a assertion and reason type question as classified in NCERT Exemplar

(A) Explanation:

Alkaline earth metals are stable at room temperature except beryllium carbonate. It decomposes to give beryllium oxide and carbon dioxide. It is kept in an atmosphere of carbon dioxide so that equilibrium shifts to the right. The reaction is shown below. BeCO₃ → BeO+CO₂

Answer-A
Only first option clearly defines the meaning of HIF, by using the term 'protein'. Option C can be a characteristic but not a complete meaning. The meaning can be derived with the help of the following lines “When oxygen levels decrease, the HIF protein inhibits oxygen-consuming processes of the cells by altering the activity of numerous genes”.

This is a long answer type question as classified in NCERT Exemplar

Alkali metals are ionic in nature due to their larger size. They have +1 oxidation states. Alkali metals forms three types of oxides such as peroxides, superoxides and normal oxides. The basic character of normal oxides increases from lithium oxide to caesium oxide.

The halides of alkali metals are also ionic except lithium halide. Lithium halide is covalent in nature because of small size and high polarizing power.

Oxosalts of alkali metal are solid water-soluble ionic compounds. Oxosalts of lithium show different properties due to small size of lithium.

This is a long answer type question as classified in NCERT Exemplar

(i) The alkali metals dissolve in liquid ammonia and give a blue solution, which is conductive in nature. A solution of sodium in liquid ammonia at -30C conducts electricity. The ammoniated electrons are responsible for the blue color of the solution as they absorb energy in the visible region of light and impart blue color to the solution. Both the ammoniated cations and ammoniated electrons are responsible for the electrical conductivity of the solution.

Na + (x+y)NH₃→ [Na (NH₃)_x]⁺ + [e (NH₃)_y]⁻

(ii) The blue color changes to bronze color in concentrated solution due to the formation of a cluster of metal ions. The standing blue solution liberates hydrogen gas with the formation of amide.

M⁺ + e^? + NH₃? MNH₂ + 12H₂

This is a long answer type question as classified in NCERT Exemplar

As the size of metal ions increases, the stability of peroxides and superoxides increases. Peroxide and superoxide ions combine with a large size of alkali metals. Lithium forms monoxide, sodium forms peroxide and potassium, rubidium and caesium forms superoxide.

Li+O₂→Li₂O

Na+O₂→Na₂O₂

 K+O₂→KO₂

This is a long answer type question as classified in NCERT Exemplar

Compound A reacts with water to form compound B. So, the compound A is calcium oxide. When water is added to calcium oxide, calcium hydroxide is formed. It is lime water. The compound gives a milky appearance which is compound C. The compound C is calcium carbonate. On passing, excess carbon dioxide milkiness disappears due to the formation of compound D. The compound is calcium hydrogen carbonate. The reactions are as follows:

CaO+H₂O→Ca (OH)₂

Ca (OH)₂+CO2→CaCO₃

Ca (OH)₂+CO₂+H2O→Ca (HCO₃)₂

This is a long answer type question as classified in NCERT Exemplar

Beryllium being least reactive does not form hydride by direct heating with dihydrogen. It is prepared by reacting it with lithium aluminum hydride. The reactions are shown below.

8LiH + Al₂Cl₆→2LiAlH₄+6LiCl

2BeCl₂+LiAlH₄→2BeH₂+LiCl+AlCl₃

This is a long answer type question as classified in NCERT Exemplar

In group 2, only beryllium is amphoteric in nature which means it reacts with both acids and bases. Also, beryllium only forms covalent oxide due to the covalent nature. So, the element is beryllium.

It reacts with acid to form beryllium chloride and it reacts with base to form beryllate ion which is soluble in sodium hydroxide. The reaction is shown below.

Be (OH)₂+2OH⁻ → [Be (OH)₄]²⁻

Be (OH)₂+2HCl →BeCl₂+2H₂O

This is a long answer type question as classified in NCERT Exemplar

The element imparts yellow colour to the flame in flame test which means the element of group 1 is sodium. Sodium is used in the transmission of nerve signals and transport of sugars and amino acids into cells. Reactions are shown below.

2Na+O₂→Na₂O₂

4Na+O₂→2Na₂O

2Na₂O+O₂→2Na₂O₂

This is a short answer type question as classified in NCERT Exemplar

Lithium has the highest negative reduction potential value. Due to the small size of lithium it has the highest ionization enthalpy. Due to this, the reducing power of lithium is the highest in an aqueous solution.

This is a short answer type question as classified in NCERT Exemplar

Alkali metals forms oxide when reacted with air. Lithium forms monoxide, sodium forms peroxide and potassium forms superoxide.

4Li+O₂→2Li₂O

Na+O₂→Na₂O₂

 K+O₂→KO₂

This is a short answer type question as classified in NCERT Exemplar

(i) O₂²⁻+ 2H₂O → H₂O₂+2OH⁻

(ii) 2O₂⁻+2H₂O→2OH⁻+H₂O₂+O₂

This is a short answer type question as classified in NCERT Exemplar

These two elements have similar properties because of their similar atomic and ionic radii.

(i) Both are lighter element and harder than the other metals in their respective groups.

(ii) The halides of both elements, LiCl and MgCl₂ are soluble in ethanol.

This is a short answer type question as classified in NCERT Exemplar

The element from group 2 is beryllium. Be (OH)₂ is amphoteric. It reacts with both acids and bases. It reacts with acid to form beryllium chloride and it reacts with base to form beryllate ion which is soluble in sodium hydroxide. The reaction is shown below.

Be (OH)₂+2OH−→ [Be (OH)₄]²⁻

Be (OH)₂+2HCl→BeCl₂+2H₂O

Beryllium sulfate are readily soluble in water.

This is a short answer type question as classified in NCERT Exemplar

(i) As the size of the cation increases, the thermal stability of carbonate increases. The more stable will be the oxide of an alkaline earth metal, the less stable will be the carbonate. Hence, beryllium carbonate will be highly unstable because its oxide will be stable.

(ii) Alkali metals and alkaline earth metals form oxides with oxygen and give metal oxides. The oxides are basic in nature. BeO is an exception because BeO is amphoteric.

They also react with water to form sparingly soluble hydroxides. On increasing the size of the cations beryllium oxide and magnesium oxide have the highest lattice energy and they are insoluble in water.

This is a short answer type question as classified in NCERT Exemplar

BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₄ and BaSO₄ are insoluble The greater hydration enthalpy of Be²⁺ and Mg²⁺ ions overcome the lattice enthalpy factor and therefore, their sulfates are soluble.

This is a short answer type question as classified in NCERT Exemplar

Ionic compounds are formed from the alkali metals due to their large ionic size and low ionization enthalpy. Thus, they are soluble in water. But, due to the small ionic size, high ionization enthalpy, and high electronegativity of lithium, it forms compounds of covalent nature and thus, are soluble in organic solvents.

This is a short answer type question as classified in NCERT Exemplar

In the Solvay process, carbon dioxide is transferred through a concentrated solution of ammoniacontaining sodium chloride, which forms ammonium carbonate followed by ammonium hydrogen carbonate. The chemicals in ammonium hydrogen carbonate are different and are heated to form sodium carbonate. NH₃ is found in a solution containing NH₄Cl that is heated and treated with Ca (OH)₂? The reaction of (NH₄)2CO₃ with NaCl provides two products, Na₂CO₃ and NH₄Cl both soluble in water which do not shift to the right balance.

This is a short answer type question as classified in NCERT Exemplar

The flame is due to the excitation of the electron from its higher energy state to the lower energy states. Due to the small atomic and ionic size of beryllium and magnesium, electrons are tightly bound to the atom. The electrons of Be and Mg does not gain excitation from the energy provided by the flame. Hence they do not show any flame in the flame test.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iv)

In alkali metals, on going down the group, as the metallic strength decreases, melting point  decreases. Thus, Cs has the lowest melting point and melts at 30°C.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (i)

The reactivity of alkali metals increases on moving down the group. So, Li is least reactive. It will react with water least vigorously.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)

Li due to its small size it has a high value of hydration enthalpy. Due to the high hydration enthalpy of Li atom, it is the strongest reducing agent in the aqueous medium.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (iv)

BaCO₃ will be the most thermal stable. The size of Ba²⁺ is very large due to which it has low Polarising power and cannot polarise the oxygen atom. As the positive ions gets larger on moving down the group, the effect on the carbonate ion near them decreases. Thus, the carbonate ions of large cations are stable.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

Due to the smaller size of the cation Be²⁺, and larger size of the anion CO₃^2? , BeCO₃ is unstable in air. Therefore, it is kept in CO₂ atmosphere to avoid decomposition.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

As the size of the metal increases, the basic character of the hydroxide increases down the group. The solubility of hydroxide increases, and hydroxide of Be and Mg is almost insoluble.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

BeCl₂ is covalent in nature due to the high polarising power of beryllium (+2) ion. On moving down the group, as the size increases, covalent nature decreases. Therefore, beryllium chloride being most covalent is soluble in ethanol.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)On moving down the group as the size increases, the nuclear charge attraction decreases. The effect of increasing size outweighs the increasing nuclear charge. Thus, ionization enthalpy decreases. The order of decreasing ionization enthalpy is,

Li>Na>K>Rb

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii)

The solubility of salt depends on the lattice and hydration energy both. By increasing lattice energy and decreasing hydration energy solubility decreases. In case of biE, both the ions are small, thus the compound has high lattice energy and is insoluble in water. If both the cations and anions are comparable in size, then stability is maximum and solubility is minimum

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

Be (OH)₂ is amphoteric. It reacts with both acids and bases. It reacts with acid to form beryllium chloride and it reacts with base to form beryllate ion which is soluble in sodium hydroxide. The reaction is shown below.

Be (OH)₂+2OH⁻→ [Be (OH)₄]²⁻

Be (OH)₂+2HCl→BeCl₂+2H₂O

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

Sodium carbonate is prepared by Solvay's process. In this process, ammonia is recovered when ammonium chloride reacts with calcium hydroxide. On reacting NH₄Cl with Ca (OH)₂ calcium chloride is obtained as a by-product. The reaction is given below.

2NH₄Cl+Ca (OH)₂→2NH₃+CaCl₂+H₂O

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

The alkali metals dissolve in liquid ammonia and give a blue solution, which is conductive in nature. A solution of sodium in liquid ammonia at −30^? C conducts electricity. The ammoniated electrons are responsible for the blue color of the solution as they absorb energy in the visible region of light and impart blue color to the solution.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii)

On adding water to cement it gets harder over a period of time. This is known as setting of cement. Gypsum is added to the cement to slow the setting time of cement which occurs in the presence of water.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

An anhydrous calcium sulfate is dead burnt paris. It is known as dead burnt because it does not set like plaster when moistened with water. When gypsum is heated the water of crystallization is lost and an anhydrous calcium sulphate is obtained.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)

Ca (OH)₂ saked lime is only sparingly soluble in water. When CaO is soaked in water, it produces a white suspension, and the reaction is highly exothermic. White suspension is called milk of lime and it is used for white washing.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

Beryllium being least reactive does not form hydride by direct heating with dihydrogen. Also due to its small size and high ionization enthalpy it cannot form hydrides by direct heating.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (iv)

Anhydrous sodium carbonate is known as soda ash. The formula of soda ash is Na₂CO₃? It is also known as washing soda.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii)

Alkali and alkaline metals impart characteristic color. Calcium gives brick red color, Strontium gives crimson red color, Barium gives apple green color, and radium gives crimson color. So, the compound is calcium nitrate. On heating calcium nitrate, it gives calcium oxide, nitrogen and nitrogen dioxide gas which is brown in color.

Ca (NO₃)²→CaO+O₂+NO₂↑

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i)

Calcium hydroxide is used in the preparation of bleaching powder. It reacts with chlorine to form hypochlorite, a constituent of bleaching powder.

Ca (OH)₂+Cl₂→CaCl₂+H₂O+Ca (OCl)₂

It is a white amorphous powder and is sparingly soluble in water.

This is a multiple choice type question as classified in NCERT Exemplar

Correct Option (iii)

The chemical formula of A is Ca (OH)₂ Sodium carbonate is prepared by Solvay's process. In this process, ammonia is recovered when ammonium chloride reacts with calcium hydroxide. On reacting NH₄Cl with Ca (OH)₂, calcium chloride is obtained as a by-product. The reaction is given below.

2NH₄Cl+Ca (OH)₂→2NH₃+CaCl₂+H₂O

When CO₂ is bubbled through an aqueous solution of Ca (OH)₂ the solution turns milky. It is used in whitewashing due to its disinfectant nature.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (iv)

Chlorides of alkaline earth metals are hydrated salts. They can be used as dehydrating agents due to their hygroscopic nature to absorb moisture from air. Extent of hydration decreases from Mg to Ba.

Therefore, dehydration of hydrates of halides of calcium, barium and strontium i.e., CaCl₂.6H₂O, BaCl₂.2H₂O, SrCl₂.2H₂O can be achieved by heating.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii) and (iv)

Alkali metals have larger atomic size and low density. These metals lose electrons due to less effective nuclear charge. They have a high value of electrode potential.

Lithium has the highest negative reduction potential value. Due to the small size of lithium it has a small atomic size and  the highest ionization enthalpy.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i) and (iii)

Na₂CO₃ is used in paper paints and textile industry. It is used to manufacture soap powders.

NaOH is used in textile industry for mercerising cotton fabrics.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i) and (ii)

On moving down the group, as the size increases the hydration enthalpy decreases. The hydration enthalpy of beryllium and magnesium ions are high due to the small size. Thus, sulfates are readily soluble in water.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (ii) and (iii)

Zeolite is sodium aluminum silicate. It is used to reduce the hardness of water. It has a property to exchange Mg²⁺ ions and Ca²⁺ ions from hard water by the Na⁺ions of zeolite.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i) and (iii)

The chlorides of alklaje earth metals are hydrated. On moving down the group, the extent of hydration decreases. Therefore, the correct formula of halides are: CaCl₂?6H₂O and BaCl₂?2H₂O.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i) and (ii)

Beryllium resembles aluminum through the diagonal relation. Beryllium forms a protective film of oxide on the surface, and thus, is prevented by the attack of acids.

Thus, statement (1) is correct.

On moving down the group, as the size increases the hydration enthalpy decreases. The hydration enthalpy of beryllium ions are high due to the small size. Thus, sulfates are readily soluble in water. Thus, statement (2) is correct.

Beryllium does not exhibit coordination numbers more than four. As it has no d orbitals.

Thus, statement (3) is not correct.

Beryllium oxide is amphoteric in nature. It reacts with both acids and bases. It reacts with acid to form beryllium chloride and it reacts with base to form becyllate ion which is soluble in sodium hydroxide. The reaction is shown below.

Be (OH)₂+2OH−→ [Be (OH)₄]²⁻

Be (OH)₂+2HCl→BeCl₂+2H₂O

Thus, statement (4) is not correct.

This is a multiple choice type question as classified in NCERT Exemplar

Correct option (i) and (ii)

Among the alkali metals, lithium has an exceptionally small size. Due to the small size of lithium and high nuclear charge it has high palajising power.

This is a matching type question as classified in NCERT Exemplar

(i) → (c); (ii) → (b); (iii) → (d) : (iv) → (a); (e)

(i) Lithium has the highest negative reduction potential value. Due to the small size of lithium it has a small atomic size, the highest ionization enthalpy Due to this, the reducing power of lithium is the highest.

(ii) Sodium is a stronger monoacidic base due to lower first ionization enthalpy of sodium. Also sodium forms sodium hydroxide  Therefore 1 mole of it replaces one mole of hydrogen ions from acids.

(iii) Calcium oxalate is an ionic compound because its hydration enthalpy is low. It is highly insoluble and dissolves poorly in water.

(iv) Barium sulfate is insoluble due to the high hydration energy. The electronic configuration of barium is 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s²

Thus, option (iv) from column I is matched with (a) and (e) from column II.

This is a matching type question as classified in NCERT Exemplar

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

(i) CaCO₃ is used in manufacturing of high quality paper.

(ii) Due to the disinfectant nature of Ca (OH)₂ is used in white wash.

(iii) CaO is used in the manufacture of sodium carbonate from caustic soda.

  (iv)CaSO₄ is used in dentistry, ornamental work and for making statues.

[Mn₂ (CO)₁₀]

Bridging ligand (CO) is (0)

$P\left(V_m-b\right)=RT$

$\frac{P{V}_m}{RT}-\frac{Pb}{RT}=1$

$Z=1+\frac{Pb}{RT}$

${\left(\frac{\partial z}{\partial P}\right)}_T=\frac{b}{RT}$

So, comparing with $\frac{xb}{RT}$

x = 1

$Mn{O}_4^{-}+8H+5e^{-}\to Mn+4H_{2}O{E}^o=1.51V$

Quantity of electricity required to reduce 1 mole of $Mn{O}_4^{-}$ is 5F

So, for 5 mole $Mn{O}_4^{-}$ 25F electricity is required.

X → Y

$\begin{array}{l}{E}_{a_f}-E_{a_b}=-20\\ 30-E_{a_b}=-20\end{array}$

$E_{a_b}=50kJ/mole$

$A{B}_2\left(g\right)?A\left(g\right)+2B\left(g\right)$

Initial1 mole -

At equilirbium 1-x mole x mole2x mole

V = 25 L and

T = 300 K.

At equilibrium, P = 1.9 atm

Total moles at equilibrium, n = 1 + 2x

V = 25 L

T = 300 K

Using, PV = nRT

1.9 × 25 = (1 + 2x) × 0.08206 × 300

x = 0.465

Now;Partial pressure of AB2 at equilibrium = $\frac{0.535}{1.93}\times 1.9atm$

Using ; $K_p=\frac{P_A.P_B^2}{P_{A{B}_2}}$

$K_p=\frac{\left(\frac{0.465\times 1.9}{1.93}\right){\left(\frac{2\times 0.465}{1.93}\times 1.9\right)}^2}{\frac{0.535}{1.93}\times 1.9}$ = 0.728

K_P = 0.73

K_p = 73 × 10^-2

So; x = 73

$C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$

Dichromate ion converted to chromate ion in basic medium and oxidation number of Cr in $Cr{O}_4^{2-}$ is +6.

PV = nRT

1 × V = $\frac{3.12}{32}\times 0.0821\times 300$

V = 2.4 litre

Vol of O₂ adsorbed per gm = 2.4 / 1.2 = 2 litre

Moles of SO₂ = $\frac{224\times 10^{-3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 × 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18}=2$

Using ; relative lowering in V.P

$\frac{P^o-P}{P^o}=i{x}_B$

Where; $\Delta P=P^0-P$ is lowering in V.P

$\Delta P=P^{0}i{x}_B$

i for NaHSO₃ = 2

here; $x_B=\frac{n_B}{n_A}$ since solution is dilute

$\Delta P=24\times 2\times \frac{0.01}{2}=0.24$

$\Delta P=24\times 10^{-2}mmHg$

So; x = 24

50000.020 * 10^-3

The significant figure in the given number is 8.

${\Delta }_r{H}^o=80kJmo{l}^{-1}$

${\Delta }_r{S}^o=2TkJmo{l}^{-1}$

For a reaction to be spontaneous;

${\Delta }_r{S}^o>\frac{{\Delta }_r{H}^o}{T}$

$T>\frac{{\Delta }_r{H}^o}{{\Delta }_r{S}^o}$

$T>\frac{80\times 10^3}{2T}$

$T^2>40000K$

$T>200K$

So, minimum T at which reaction will be spontaneous is 200 K.

$\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{K}}_2{\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7\stackrel{{\mathrm{H}}^{+}}{?}{\mathrm{C}\mathrm{r}\mathrm{O}}_2{\mathrm{C}\mathrm{l}}_2\stackrel{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}{?}{\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4\underset{{\mathrm{H}}^{+}}{\stackrel{{\mathrm{H}}_2{\mathrm{O}}_2}{?}\underset{\left(\mathrm{C}\right)}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5}}$

 ${\mathrm{C}\mathrm{r}\mathrm{O}}_2{\mathrm{C}\mathrm{l}}_2\to 5$

${\mathrm{N}\mathrm{a}}_2{\mathrm{C}\mathrm{r}\mathrm{O}}_4\to 7$

${\mathrm{C}\mathrm{r}\mathrm{O}}_3\to 6$

192.5

$\mathrm{}2\mathrm{C}\left(\mathrm{g}\mathrm{r}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)?{\mathrm{C}}_2{\mathrm{H}}_6\left(\mathrm{g}\right)$

$\mathrm{\Delta }{\mathrm{H}}_{\mathrm{f}}=2\times \mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left(\mathrm{C}\right)+3\times \mathrm{\Delta }\sigma {\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left({\mathrm{H}}_2\right)-\mathrm{\Delta }{\mathrm{H}}_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{b}}\left({\mathrm{C}}_2{\mathrm{H}}_6\right)$

$\Rightarrow \sigma {\mathrm{H}}_{\mathrm{f}}=2(-286)+3(-393.5)-(-1560)=192.5$

$\mathrm{}0.3675$

$30\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$ of $\mathrm{H}\mathrm{C}\mathrm{l}$

Then in $250\mathrm{m}\mathrm{L}\mathrm{H}\mathrm{C}\mathrm{l}$ will be $\frac{30}{4}\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

${\mathrm{H}}_2{\mathrm{S}\mathrm{O}}_4$ will be $\frac{1}{2}\times \frac{30}{4}\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}=98\times \frac{1}{2}\times \frac{30}{4}\times {16}^{-3}\mathrm{g}=0.3675\mathrm{g}$

The dotted ones are ${\mathrm{s}\mathrm{p}}^2$  carbons

$500\mathrm{m}\mathrm{L}$ has $\mathrm{H}\mathrm{C}\mathrm{l}=25\times {10}^{-3}\mathrm{m}\mathrm{o}\mathrm{l}=25\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$120\mathrm{m}\mathrm{L}$ has $=1\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

and $500\mathrm{m}\mathrm{L}$ has ${\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}=\frac{1}{20}\times {10}^3\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

so $20\mathrm{m}\mathrm{L}$ has $\frac{{10}^3}{20}\times \frac{20}{500}=2\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ added in $20\mathrm{m}\mathrm{L}$ is $5\times \frac{1}{2}=2.5\mathrm{m}\cdot \mathrm{m}\mathrm{o}\mathrm{l}$

So, $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}$ (left) $+{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}?{\mathrm{C}\mathrm{H}}_3\mathrm{C}\mathrm{O}\mathrm{O}\mathrm{H}+{\mathrm{H}}_2\mathrm{O}$
1.5
2
1.5

$\Rightarrow \mathrm{p}\mathrm{H}={\mathrm{p}}^{{\mathrm{K}}_{\mathrm{a}}}+\mathrm{l}\mathrm{o}\mathrm{g}?\frac{\text{salt}}{\text{acid}}=4.75+\mathrm{l}\mathrm{o}\mathrm{g}?0.4771=5.2271$