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

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

The given reaction is a redox change. 

2Na (s)+H₂ (g)→2NaH (s)   

The half reaction is:

2Na (s)→2Na⁺ (g)+2e^-

The other half reaction is:

H₂ (g)+2e^-→2H^- (g)

This splitting of the reaction into two half-reactions automatically reveals here that sodium is oxidized, and hydrogen is reduced. Any substance which loses electron is oxidized and gains electron is reduced hence is the case of sodium and hydrogen atoms respectively. Hence, the complete reaction is a redox change.

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

On the basis of standard reduction potential suggested in the reactivity series (ii) reaction can take place as Mg has more negative value of E? cell. Hence, Mg will be oxidized by losing electron and iron will be reduced by gaining electron.

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

Disproportionate is defined as the reaction in which one compound of intermediate oxidation state converts to two compounds, one of higher and one of lower oxidation states So, to occur such type of redox reaction, the element should exist in at least three oxidation states. So that element present in the intermediate state and it can change to both higher and lower oxidation state during disproportionate reaction. Fluorine is the most electronegative element and a strong oxidizing agent and is the smallest in size of all the halogens. It does not show a positive oxidation state (shows only −1 oxidation state) and hence, does not undergo disproportionate reaction.

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

Redox couple are given as

Cu²⁺/Cu and Zn²⁺/Zn

Mg²⁺/Mg and Fe²⁺ /Fe

Br₂/Br^- and Cl₂ / Cl^-

Fe²⁺ /Fe and Cd²⁺/Cd

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

We can calculate the oxidation states by - ]

NaClO₄ Oxidation no. of chlorine = +7 

Suppose oxidation number of chlorine is x then,  1 + x + 4 × (−2) = 0

∴ x - 7 = 0 

    x = +7

We can calculate, the oxidation states, as given below-

NaClO₃ Oxidation no. of chlorine = +5

? NaClO Oxidation no. of chlorine = +1

? KClO₂ Oxidation no. of chlorine = +3

? Cl₂O₇ Oxidation no. of chlorine = +7

ClO₃ Oxidation no. of chlorine = +6

Cl₂O Oxidation no. of chlorine = +1 

NaCl Oxidation no. of chlorine = −1

Cl₂ Oxidation no. of chlorine = 0

ClO₂ Oxidation no. of chlorine = +4.

Oxidation state (+2) is not present in any of the above compounds.

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

We can Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant.  Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant. Example Measurement of standard electrode potential of electrode E? _Zn2+/Zn using SHE as a reference electrode.

Types of Redox Reactions

The different types of Redox Reactions are:

Decomposition Reaction

Combination Reaction

Displacement Reaction

Disproportionation Reactions

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

The given reaction is as below-

Cl₂  (g) +2OH^--  (aq)→ClO^-- (aq)+ Cl^-- (aq) + H₂O  (l)

In the given reaction, oxidation number  of Cl increases from 0 (in Cl₂) to +1 (in ClO^-) and decreases to -1 (in Cl^-). Therefore, Cl₂ is both oxidized to ClO^- and reduced to Cl^-. Since Cl^- ion cannot act as an oxidizing agent (because it cannot decrease its O.N. lower than -1), hence, Cl₂ bleaches substances due to oxidizing action of hypochlorite, ClO ion.

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

In MnO₄^2–, Mn is in the highest oxidation state that is +7 hence here manganese cannot undergo oxidation that is why disproportionate is not possible whereas in  MnO₄^2- manganese is in +6 oxidation state which can be oxidized as well as reduced.

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

The given compound can differ in reactivity as-Lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO₂ thus, can act as an oxidant (oxidising agent) and, therefore, can oxidise chloride ions of HCl into chlorine.

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

Since nitric acid itself is an oxidising agent therefore, it is unlikely that the reaction may occur between PbO₂ and nitric acid. However, the acid-base reaction occurs between PbO and nitric acid because PbO is a basic oxide.

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

(i) 2Mn0^–₄ + 5S0₂ + 2H₂0 + H⁺?5HS0^–₄ + 2Mn²⁺
Balancing by ion-electron method:

(ii) We can balance the given reaction by oxidation number method-

Balancing by oxidation number method as to make the electron gain and loss equal as given

3N₂H₄ + 4ClO₃^- → 6NO + 4Cl^- + 6H₂O

The balanced chemical is given as-

(iii) We can balance the given reaction by ion electron method as

Cl₂O₇(g) + H₂O₂ (aq) → ClO₂^- +O₂ (acidic medium)

Balancing bu ion electron method

2 × { Cl₂O₇ + 6H⁺ + 8e^- → 2ClO₂^- + 3H₂O

8  × { H₂O₂ → O₂ + 2H⁺ + 2e^- 

The balanced chemical is given as-

2Cl₂O₇ + 12H⁺ + 16e^- → 4ClO₂^- + 6H₂O

                         8H₂O₂ → 8O₂ + 16H⁺ + 16e^-

          2Cl₂O₇ + 8H₂O₂ → 4ClO₂^- + 6H₂O + 8O₂ + 4H⁺

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

(i) Let the oxidation number of P in HPO₃^2- be x. 

Therefore, +1 + x + (-6) = -2 

x = +3

Oxidation number of phosphorus is= +3.

(ii) Let the oxidation number of P in PO₄^3- be x. 

Therefore, x + (-8) = -3

x = +5 

Oxidation number of phosphorus is= +5.

(a) let x= oxidation number of   sulphur, and   +1 is oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + 2x - 6 = 0 

x = +2. 

(b) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(c) let x= oxidation number of   sulphur, and   +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 6 = 0

x = +4

(d)  let x= oxidation number of sulphur, and  +1 is   oxidation number of Na, -2 is oxidation number of O, also we can assume total charge on compound = 0 then solving we get.

+2 + x - 8=0

x = +6.

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

(i) We can balance the given equation by oxidation number method- 

(a)Balance the increase and decrease in O.N.

  (b) Balancing H and O atoms by adding H⁺ and H₂O molecules

 

(ii) We can balance the given equation by oxidation number method- 

Total decrease in O.N. = 1

To equilize O.N. multiply NO₃^-, by 10

I₂ + 10 NO₃^-? 10NO₂ + IO₃^-

Balancing atoms other than O and H

I₂ + 10 NO₃^-? 10NO₂ + 2 IO₃^-

Balancing O and H

I₂ + 10 NO₃^- + 8H⁺? 10NO₂ + 2 IO₃^- + 4H₂O

 

(iii) We can balance the given equation by oxidation number method

Total increase in O.N. = 0.5 × 4 = 2

Total increase in O.N. = 1 × 2 = 2

to equilize O.N. multiply S₂O₃^2– and I^- by 2.

 

(iv) We can balance the given equation by oxidation number method-

The balanced chemical reaction is given as-

Total increase in O.N. = 5 × 2= 10

To equalize O.N. multiply CO₂ by 2. 

MnO₂   + C₂O₄^2-   ?  Mn₂⁺   + 2CO₂

Balance H and O by adding 2H₂O   on right side, and 4H+  on left side of equation.

MnO₂   + C₂O₄^2-   + 4H⁺   ?  Mn₂⁺   + 2CO₂   + 2H₂O .

 

(i) We can write the given reaction along with their oxidation numbers as-

As, chlorine is oxidised in hydrochloric acid (behaving as reducing agent) -as its oxidation number is increases during the reaction from -1 to 0) and nitric acid is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +5 to +3). Hence, it is the redox reaction.

(ii) We can write the given reaction along with their oxidation numbers as-

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

(iii) We can write the given reaction along with their oxidation numbers as-

As, carbon is oxidised in carbon monoxide (behaving as reducing agent) -as its oxidation number is increases during the reaction from +2 to +4) and ferric oxide is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from +3to 0). Hence, it is the redox reaction.

(iv) We can write the given reaction along with their oxidation numbers as-

Here, oxidation number of none of the atoms change hence it is not a redox reaction.

(v) We can write the given reaction along with their oxidation numbers as-

As, nitrogen is oxidised in ammonia (behaving as reducing agent) -as its oxidation number is increases during the reaction from -3 to 0) and oxygen is reduced (behaving as oxidising agent) as its oxidation number is decreases during the reaction from 0 to -2). Hence, it is the redox reaction.

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

(i) The balanced chemical is given as –

 

(ii) The balanced chemical is given as –

 

(iii) The balanced chemical is given as –

 

(iv) The balanced chemical is given as –

 

 

 

 

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iv) is the correct answer.

Redox reaction is defined as the simultaneous oxidation and reduction of reacting species. Thus, change in oxidation state will decide whether a reaction is redox or not. Thus, assigning the oxidation states as:

(i) CuO + H₂→ Cu + H₂O

 Here, oxidation of H and reduction of Cu is taking place.

(ii) Fe2O₃ + 3CO→ 2Fe + 3CO₂

Here, oxidation of C and reduction of Fe is taking place.

(iii) 2K + F₂→2KF

Here, the oxidation of K and reduction of F is taking place.

(iv) BaCl₂ + H₂SO₄→ BaSO₄ + 2HCl

This is not a redox reaction, but it is a double displacement reaction.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer

The strongest oxidising agent means it has greater tendency to oxidise other species and itself gets easily reduced. So higher the E^? values, stronger is the oxidising agent it is. Thus, Ag⁺ having the highest positive E^?   value among the given systems, is the strongest oxidising agent.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer

Reduction potential is defined as the tendency of the specie to get reduced. More positive the value of E® , greater is the tendency of the species to get reduced and stronger is the oxidising agent. 

On the basis of the given E®  values, the order of getting reduced is:

Br₂Ag + I₂Cu²⁺

Hence, Cu has the least tendency to get reduced and will itself gets oxidise and reduce other species as: Br₂? , Ag⁺ and I₂.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer.

E? will be negative for the pair Ag and Fe³⁺. Hence the reaction is not feasible.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) is the correct answer.

Standard reduction potential of bromine is higher than Iodine, hence Bromine is a stronger oxidant than iodine.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) is the correct answer

In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii) is the correct answer.

The oxidation states are given below-

In NH₂OH oxidation of N is -1.

NH₄NO₃ exists as NH₄⁺.NO₃^-,    

Thus, the oxidation state of N in NH₄⁺ is -3 while in NO₃^? is +5.

In N₂H₄ oxidation of N is -2

In N₃H oxidation of N is $\frac{-1}{3}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) is the correct answer

Let the oxidation number of central atom be 'y'.

Oxidation number of O = −2

CrO₂⁻: y + 2 × (−2) = −1

y − 4 = −1 or y = +3

CrO₄²⁻? : y + 4 × (−2) = −2

y − 8 = −2 or y = +6

ClO₃⁻: y + 3 × (−2) = −1

y − 6 = −1 or y = +5

MnO₄⁻: y + 4 × (−2) = −1

y − 8 = −1 or y = +7

increasing order of oxidation number of the central atom is:

CrO₂ ^– , ClO₃ ^–, CrO₄ ^2–, MnO₄

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer

For 3d¹ 4s² can exhibit the highest oxidation state as 2+1 = +3,

For 3d³4s²can exhibit the highest oxidation state as 2+3 = +5

For 3d⁵4s¹can exhibit the highest oxidation state as 1+5 = +6

For 3d⁵4s²can exhibit the highest oxidation state as 2+5 = +7

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iv) is the correct answer.

Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-

2NO₂ + 2OH^- →NO₂ ^-+ NO₃^- +H₂O

In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in NO₃⁻?  to +5 in NO₂ ? and decreases from +4 in NO to +3 in NO₂⁻.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii) is the correct answer.

As, Cl, Br, I all are having -1 to +7 oxidation state. But Oxidation state of F is fixed (-1) as it is the most electronegative element and do not loose electron. Hence, it does not show disproportionate tendency.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Options (iii) and (iv) are the correct answers.

As, in the given reaction below-

2KClO₃→2KCl+3O₂

Potassium remains in same oxidation state and oxygen is being oxidized

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii) and (iv) is the correct answer.

The given equation is as-

Zn+2HCl→ZnCl₂+H₂

In this Zn has a negative E? value, which means it will undergo oxidation and will act as a reducing agent (reductant). Zn can produce H₂ gas with HCl, as hydrogen has higher standard reduction potential than Zn, hydrogen will undergo reduction and will act as oxidant.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii), (iii) and (iv) are the correct answers.

Elements that are having only s-electrons in the valence shell do not show more than one oxidation state (shows only one oxidation state of +1).  Hence, (b), (c) having incompletely filled d-orbital's in the outermost shell show variable oxidation states.  Element with outer electronic configuration as 3d¹ 4S² shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d²4S² shows variable oxidation states of +2, +3 and +4.  P-Block elements also show variable oxidation states due to involvement of d-orbital's and inert pair effect. Hence, element having 3S² 3P³ as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbital's.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Hence, option (iii) and (iv) are the correct answers.

The given reaction is as below-

P₄+3OH^- +3H₂O→PH₃ + 3H₂PO₂^-

The above reaction is a kind of disproportionate reaction in which phosphorous is being reduced as well as oxidized whereas hydrogen remains same in +1 oxidation state.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) and (ii) are the correct answers.

The ones which will act as anodes when connected to standard hydrogen electrode as they have more negative standard reduction potential as compared to standard hydrogen electrode. The one which will act as cathodes when connected to standard hydrogen electrode as they have more positive standard reduction potential as compared to standard hydrogen electroDE

This is a Matching Type Questions as classified in NCERT Exemplar

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

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f).

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

Option (i) is the correct answer.

Fluorine is most electronegative element that is why it is best oxidant among halogens

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

Option (iii) is the correct answer

As permanganate ion changes to MnO₂

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

Option (i) is the correct answer.

Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O₂ undergoing oxidation and decreases to -2 oxidation state in H₂O undergoing reduction

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

Option (ii) is the correct answer.

A redox couple is defined as pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half reaction.

Gabriel phthalimide synthesis is used for 1° Aliphatic /alicyclic amine

1° Amine

CH₃ – CH₂ – NH₂

$C_{12}{H}_{22}{O}_{11}+H_{2}O\to \underset{Glucose}{C_6{H}_{12}{O}_6}+\underset{Fructose}{C_6{H}_{12}{O}_6}$

$K=\frac{2.303}{t}log\frac{a}{a-x}$

$\frac{kt}{2.303}=log\frac{a}{a-x}$

$\frac{ln2\times 9}{\frac{10}{3}\times 2.303}=log\left(\frac{1}{f}\right)$

$log\left(\frac{1}{f}\right)=81.24\times 10^{-2}=81\times 10^{-2}$

$Pb{l}_2?P{b}^{2+}+2l^{-}{K}_{sq}=8.0\times 10^{-9}$

Pb (NO₃)₂ -> Pb²⁺ + ^{$2N{O}_3^{-}$}

0.1 M-

-0.1 M    0.1 M

  $K_{sq}=8\times 10^{-9}Using{K}_{sq}=\left[P{b}^{2+}\right]{\left[I^{-}\right]}^2$          

$8\times 10^{-9}=0.1\times {\left(2S\right)}^2$

S = 141 × 10^-6 M

$C_x{H}_y\left(g\right)+\left(x+\frac{y}{4}\right)O_2\left(g\right)\to xC{O}_2\left(g\right)+\frac{y}{2}{H}_{2}O\left(\mathcal{l}\right)$

V 6V - -

- - 4V

Volume of CO2 = 4 × $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

Vx = 4V

x = 4

Volume of O2 = 6 × $V_{C_x{H}_y}$ ${}_{{}_{}{}_{}}$

$V\left(x+\frac{y}{4}\right)=6V$

$4+\frac{y}{4}=6$

y = 8

$\frac{1.86}{93}=\frac{w}{135}$

W = 2.70 g

Mass produced actual = 2.70 $\times \frac{90}{100}=2.43=243\times 10^{-2}g$

W = 4.75 g

$n=\frac{4.75}{26}$ T = 323 K, R = 0.0826

  $P=\frac{740}{760}atm$          

 V = $\frac{nRT}{P}=\frac{4.75\times 0.0826\times 323}{26\left(\frac{740}{760}\right)}=5litre$

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

$E_1=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}{\left[\frac{1}{H^{+}}\right]}^8$  

[H⁺] = 1M

$E_2=E^0-\frac{0.059}{5}log\frac{\left[M{n}^{2+}\right]}{\left[Mn{O}_4^{-}\right]}+\frac{0.059}{5}log10^{-32}$   

So, difference in E₁ & E₂ is

= 0.3776 V

= 3776 × 10^-4 V

$3CH\equiv CH\left(g\right)\to C_6{H}_6\left(\mathcal{l}\right)$       

$\Delta G^0=-RT\mathcal{l}nK.........\left(i\right)$

$\Delta G^0=\sum \Delta {G^0}_P-\sum \Delta {G^0}_R..........\left(ii\right)$    

Equating (i) & (ii)

-2.303 RTlogk = 4.88 × 10⁵

$logK=\frac{-4.88\times 10^5}{2.303\times R\times T}=\frac{-488000}{5705.848}=-85.52=-855\times 10^{-1}$

So; magnitude of log K = 855 × 10^-1

  $\Delta T_f=k_f.m$

$T_f^0-T_f=5.12\times \frac{\left(\frac{10}{58}\right)}{\left(\frac{200}{1000}\right)}$

    5.5 – T_f = $\frac{5.12\times 5\times 10}{58}$  

  $T_f=1.086°C=\left(1.086\right)°C\cong 1°C$             

α - sulphur & β - sulphur – Diamagnetic, S₂ – form is paramagnetic due to presence of unpaired electron in π* orbital like O₂.

Beo & Be (OH)₂ are amphoteric in nature, because they react with both acid and base

Stability constant are :

K₁ = 10⁴

K₂ = 1.58 × 10³

K₃ = 5 × 10²

K₄ = 10²

Overall stability constant K will be

K = K₁ × K₂ × K₃ × K₄

 = 7.9 × 10¹¹

Now, overall equilibrium constant for dissociation of [Cu (NH₃)₄]²⁺ is

$K\text{'}=\frac{1}{K}=\frac{1}{7.9\times 10^{11}}=0.126\times 10^{-11}$  

= 1.26 × 10^-12

So; x = 1 (Rounded off to the nearest integer)

A (g) -> B (g)        K_P = 100

$\Delta G$ at 300 K and 1 atm

Using

$\Delta G=-RTlnK{}_P$           

$\Delta G=-R\times 300ln100$            

=-R × 300 × 2 × 2.3

$\Delta G=-1380R$            

So; x = 1380.

Initially ->              1 mol                   -

At eq.                      1-x mol               2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x        x = 1/3

Moles of Cl₂ at equibrium =   $\frac{2}{3}$

Moles of Cl at equilibrium =   $\frac{2}{3}$

Total moles =   $\frac{4}{3}$

Now: $P_{C{l}_2}=\frac{2}{4}\times 1atm$  

$=\frac{1}{2}atm$                    

$P_{Cl}=\frac{2}{4}\times 1atm=\frac{1}{2}atm$        

$K_P=\frac{{\left(P_{Cl}\right)}^2}{P_{C{l}_2}}=\frac{{\left(\frac{1}{2}\right)}^2}{\frac{1}{2}}=\frac{1}{2}$          

$K_P=0.5=5\times 10^{-1}$           

Here : x = 5

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

  $\Delta T_f=i{k}_f.m$           

  $0.5=i\times 1.86\times \frac{9.45}{94.5\times 500}\times 1000$         

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

a =  $\frac{1.344-1}{2-1}=0.344$  

Using

$K_a=\frac{C{\alpha }^2}{1-\alpha }$  

$K_a=\frac{0.2\times {\left(0.344\right)}^3}{0.66}=36\times 10^{-3}$       

So; x = 36

The balanced equation is :

$S_8\left(s\right)+12O{H}^{-}\left(aq\right)\to 4S^{2-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)+6H_{2}O\left(\mathcal{l}\right)$           

Here ; a = 12

W = 4.5 g

M.W = 90

V = 250 ml

Using M = $\frac{W}{M.W\times V}\times 1000$  

$M=\frac{4.5}{90\times 250}\times 1000=0.2$  

M = 2 × 10^-1 M

So, x = 2

Using $\lambda =\frac{h}{\sqrt[]{2qVm}}$  

${\lambda }_{Li}=\frac{h}{\sqrt[]{2\times 3e\times V\times 8.3m}}$                 m = mass of proton

${\lambda }_p=\frac{h}{\sqrt[]{2\times e\times V\times m}}$  

$\frac{{\lambda }_{Li}}{{\lambda }_p}=\sqrt[]{\frac{2eVm}{2\times 24.9eVm}}$            

= 0.2 = 2 × 10^-1

So; x = 2

Coordination no. of an atom in BCC is 8.

K = 3.3 × 10^-4 s^-1

Time for 40% completion ; t

Using $K=\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{\left[R\right]}$  

3.3 × 10^-4 =   $\frac{2.303}{t}lo{g}_{10}\frac{{\left[R\right]}_0}{0.6{\left[R\right]}_0}$

$t=\frac{2.303}{3.3}\times 10^4\times 0.22\Rightarrow t=25.58mins$  

so; the nearest integer is 26.

Phosphinic acid is hypo phosphorous acid $\left({\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2\right)$ .

$\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+{\mathrm{H}}_3{\mathrm{P}\mathrm{O}}_2?{\mathrm{N}\mathrm{a}\mathrm{H}}_2{\mathrm{P}\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

For neutrization

${\left({\mathrm{N}}_1{\mathrm{V}}_1\right)}_{\text{acid}}={\left({\mathrm{N}}_2{\mathrm{V}}_2\right)}_{\text{base}}$

$0.1*10=0.1*{\left({\mathrm{V}}_{\mathrm{m}\mathrm{L}}\right)}_{\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}}$

Asp - Glu - Lys pripeptide is:

No. of $\mathrm{C}\mathrm{O}$ group $=5$

For isotonic solution

${\mathrm{i}}_1{\mathrm{C}}_1={\mathrm{i}}_2{\mathrm{C}}_2\mathrm{}$ ffor protein $\left.\mathrm{i}=1\right\}$

${\mathrm{C}}_1={\mathrm{C}}_2$

$\frac{0.73\times 1000}{M_A\times 250}=\frac{1.65}{M_B\times 1}$

$\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}=\frac{0.73\times 4}{1.65}=1.77=177\times {10}^{-2}$

According to Faraday law

  $\mathrm{W}=\mathrm{Z}\mathrm{I}\mathrm{t}\times \eta \mathrm{}$ Where $\eta =$ efficiency

or $W=\frac{E}{96500}\times 1\times t\times \eta$

Putting values

$.104=\frac{(52/3)\times 2\times 8\times 60\times \eta }{96500}\Rightarrow \eta =0.6$

So percentage efficiency $=60\mathrm{\%}$

Number of mole of $\mathrm{X}=\frac{6.022\times {10}^{22}}{6.022\times {10}^{23}}=\frac{10}{\text{Molar mass of}\mathrm{X}}$

So molar mass of $X=100\mathrm{g}$

Molarity $=\frac{5}{100\times 2}=0.025\mathrm{M}$

Ans. $=0.025\mathrm{M}$

$\mathrm{M}=25\times {10}^{-3}$

So $P=25$

(x/m) = k (P)¹/?
log (x/m) = log k + (1/n) log P
Slope = 1/n = 2 So n = ½
Intercept ⇒ log k = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3 [0.04]² = 48 × 10?

According to IUPAC convention for naming of elements with atomic number more than 100, different digits are written in order and at the end ium is added. For digits following naming is used.

0 -nil

1-un

2-bi

3-tri

and so on.

log (k? /k? ) = (Ea / 2.303R) [1/T? - 1/T? ]
log (3.555) = (Ea / (2.303R) [1/303 - 1/313]
1.268 × 8.314 × 303 × 313 = 10Ea
So, Ea = 100 kJ