TS EAMCET 2008 Chemistry Question Paper with Answer and Solution

(C) Acetic acid $(CH_3COOH)$ on reduction with lithium aluminium hydride $(LiAlH_4)$ yields ethyl alcohol $(CH_3CH_2OH)$.
$CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH$
Acetic acid on reduction with $HI$ and red $P$ yields ethane $(CH_3-CH_3)$.
$CH_3COOH \xrightarrow{\text{Red } P + HI} CH_3-CH_3$
Therefore,reagent $A$ is $LiAlH_4$ and reagent $B$ is $HI +$ red $P$.

(D) The reaction of hydrogen halides $(HX)$ with ethyl alcohol $(C_2H_5OH)$ involves the cleavage of the $C-O$ bond,which is facilitated by the nucleophilic attack of the halide ion $(X^-)$.
As the size of the halide ion increases from $F^-$ to $I^-$,the bond dissociation energy of the $H-X$ bond decreases,and the nucleophilicity of the halide ion increases.
Therefore,the reactivity of hydrogen halides towards ethyl alcohol follows the order: $HI > HBr > HCl > HF$.

(B) The reaction of acetone $(CH_3COCH_3)$ with methyl magnesium bromide $(CH_3MgBr)$ is a nucleophilic addition reaction.
Acetone reacts with $CH_3MgBr$ to form an addition complex,$(CH_3)_3COMgBr$.
This complex,upon acid hydrolysis (decomposition with acid),yields a tertiary alcohol,$X$,which is $2$-methylpropan-$2$-ol or tert-butyl alcohol,$(CH_3)_3COH$,along with $Mg(OH)Br$.
The reaction is: $(CH_3)_2C=O + CH_3MgBr$ $\rightarrow (CH_3)_3COMgBr$ $\xrightarrow{H_3O^+} (CH_3)_3COH + Mg(OH)Br$.

(A) For an ideal transformer,the relationship between the number of turns $(N)$ and current $(I)$ is given by the inverse ratio:
$\frac{N_P}{N_S} = \frac{I_S}{I_P}$
Given:
$N_P = 50$
$N_S = 200$
$I_P = 4 ~A$
Substituting the values into the formula:
$\frac{50}{200} = \frac{I_S}{4}$
$\frac{1}{4} = \frac{I_S}{4}$
$I_S = 1 ~A$
Therefore,the current in the secondary coil is $1 ~A$.

(C) The reduction of nitrobenzene with lithium aluminium hydride $(LiAlH_4)$ in an ether solvent typically leads to the formation of azobenzene as the major product. The reaction involves the coupling of two nitrobenzene molecules.
$2C_6H_5NO_2 \xrightarrow{LiAlH_4} C_6H_5-N=N-C_6H_5$ (Azobenzene)

(A) Sucrose undergoes hydrolysis in the presence of dilute aqueous sulphuric acid to produce an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
$C_{12}H_{22}O_{11} + H_2O$ $\xrightarrow{H_2SO_4} C_6H_{12}O_6 (D-(+)\text{-glucose}) + C_6H_{12}O_6 (D-(-)\text{-fructose})$
This process results in a $1: 1$ molar ratio of the two monosaccharides.
Since the specific rotation of $D-(-)$-fructose $(-92.4^{\circ})$ is greater in magnitude than that of $D-(+)$-glucose $(+52.7^{\circ})$,the resulting mixture is laevorotatory,which is why the process is often called the inversion of cane sugar.

(B) The structure of $ClO_4^{-}$ involves one $Cl-O$ single bond and three $Cl=O$ double bonds. Each $Cl=O$ double bond consists of one $\sigma$ bond and one $d\pi - p\pi$ back bond. Thus,$ClO_4^{-}$ contains $3 \ d\pi - p\pi$ bonds.
In $XeO_3$,the xenon atom is $sp^3$ hybridized with one lone pair. It has three $Xe=O$ double bonds. Each $Xe=O$ double bond involves one $d\pi - p\pi$ bond. Therefore,$XeO_3$ contains $3 \ d\pi - p\pi$ bonds,which is equal to the number present in $ClO_4^{-}$.

(D) Given:
Observed dipole moment $= 1.03 \ D$
Bond length of $HCl$ molecule,$d = 1.275 \ \mathring{A} = 1.275 \times 10^{-8} \ cm$
Charge of electron,$e = 4.8 \times 10^{-10} \ esu$
Theoretical dipole moment $= e \times d = (4.8 \times 10^{-10} \ esu) \times (1.275 \times 10^{-8} \ cm) = 6.12 \times 10^{-18} \ esu \cdot cm = 6.12 \ D$
Percentage ionic character $= \frac{\text{Observed dipole moment}}{\text{Theoretical dipole moment}} \times 100$
Percentage ionic character $= \frac{1.03}{6.12} \times 100 = 16.83 \%$

(B) The decomposition reaction is: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$
Initial moles: $5, 0, 0$
Moles at equilibrium: $5(1-\alpha), 5\alpha, 5\alpha$
Given $\alpha = 40 \% = 0.4$,so moles at equilibrium are: $5(1-0.4) = 3$,$5(0.4) = 2$,$5(0.4) = 2$
Volume of flask $V = 500 \ mL = 0.5 \ L$
Concentrations at equilibrium: $[PCl_5] = \frac{3}{0.5} = 6 \ M$,$[PCl_3] = \frac{2}{0.5} = 4 \ M$,$[Cl_2] = \frac{2}{0.5} = 4 \ M$
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{4 \times 4}{6} = \frac{16}{6} = 2.66 \ mol \ L^{-1}$

(C) From the energy profile diagram,we have:
$E_a = \text{activation energy of forward reaction}$
$E_a^{\prime} = \text{activation energy of backward reaction}$
$E_t = \text{threshold energy}$
$1$. The diagram shows that the peak energy $(E_t)$ is higher than the energy of the reactants $(E_R)$ and products $(E_P)$.
$2$. The activation energy for the forward reaction is $E_a = E_t - E_R$.
$3$. The activation energy for the backward reaction is $E_a^{\prime} = E_t - E_P$.
$4$. Since $E_P > E_R$,it follows that $E_a > E_a^{\prime}$. Thus,option $A$ is correct.
$5$. Since the potential energy of the product is greater than that of the reactant $(E_P > E_R)$,the reaction is endothermic. Thus,option $B$ is correct.
$6$. The threshold energy $(E_t)$ is the minimum energy required for the reaction to occur,which is always greater than or equal to the activation energy of the forward reaction $(E_t = E_a + E_R)$. Therefore,the statement that threshold energy is less than the activation energy is incorrect. Thus,option $C$ is wrong.
$7$. The energy of the forward reaction is $E_a = \Delta H + E_a^{\prime}$,where $\Delta H$ is the heat of reaction. Thus,option $D$ is correct.

(C) Given,atomic number of $B = Z$. Since $B$ is a noble gas,it belongs to group $18$.
The atomic number of $A = Z-1$,which corresponds to a halogen (group $17$).
The atomic number of $C = Z+1$,which corresponds to an alkali metal (group $1$).
The atomic number of $D = Z+2$,which corresponds to an alkaline earth metal (group $2$).
Statement $(1)$: Halogens $(A)$ have high electron affinity. This is correct.
Statement $(2)$: $C$ is an alkali metal,so it exists in $+1$ oxidation state,not $+2$. This is incorrect.
Statement $(3)$: $D$ is an alkaline earth metal (group $2$). This is correct.
Therefore,statements $(1)$ and $(3)$ are correct.

(B) Let $E$ be the $EMF$ of the battery and $r$ be its internal resistance.
The terminal voltage $V$ is given by $V = E - Ir$,where $I = \frac{V}{R}$.
Thus,$E = V + \frac{V}{R}r = V(1 + \frac{r}{R})$.
For the first case: $R_1 = 16 \ \Omega$,$V_1 = 12 \ V$.
$E = 12(1 + \frac{r}{16}) = 12 + \frac{12r}{16} = 12 + \frac{3r}{4} \quad \dots (i)$
For the second case: $R_2 = 10 \ \Omega$,$V_2 = 11 \ V$.
$E = 11(1 + \frac{r}{10}) = 11 + \frac{11r}{10} \quad \dots (ii)$
Equating $(i)$ and $(ii)$:
$12 + \frac{3r}{4} = 11 + \frac{11r}{10}$
$1 = \frac{11r}{10} - \frac{3r}{4}$
$1 = \frac{22r - 15r}{20} = \frac{7r}{20}$
$r = \frac{20}{7} \ \Omega$.

(C) The circuit consists of two parallel branches $PRQ$ and $PSQ$ connected between points $P$ and $Q$.
Branch $PRQ$ has a total resistance of $3 \, \Omega + 7 \, \Omega = 10 \, \Omega$.
Branch $PSQ$ has a total resistance of $7 \, \Omega + 3 \, \Omega = 10 \, \Omega$.
Since the resistances of both branches are equal, the total current of $2 \, A$ splits equally into $1 \, A$ through each branch.
For branch $PRQ$, the current $I_1 = 1 \, A$. The potential drop across the $3 \, \Omega$ resistor is $V_P - V_R = I_1 \times 3 \, \Omega = 1 \, A \times 3 \, \Omega = 3 \, V$.
For branch $PSQ$, the current $I_2 = 1 \, A$. The potential drop across the $7 \, \Omega$ resistor is $V_P - V_S = I_2 \times 7 \, \Omega = 1 \, A \times 7 \, \Omega = 7 \, V$.
We want to find $V_R - V_S$. From the above equations:
$V_R = V_P - 3$
$V_S = V_P - 7$
Subtracting these gives: $V_R - V_S = (V_P - 3) - (V_P - 7) = -3 + 7 = +4 \, V$.

(B) Let the total current in the circuit be $I$ and the current through the galvanometer be $I_g$.
Given that $I_g = 5 \% \text{ of } I = 0.05 I$.
The shunt resistance $S$ is connected in parallel to the galvanometer $G$.
Using the current division rule,the current through the galvanometer is given by $I_g = I \times \frac{S}{S+G}$.
Substituting the values: $0.05 I = I \times \frac{S}{S+G}$.
$0.05 = \frac{S}{S+G} \implies 0.05(S+G) = S$.
$0.05 S + 0.05 G = S$.
$0.05 G = 0.95 S$.
$S = \frac{0.05 G}{0.95} = \frac{G}{19}$.

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.
Given, $\lambda = 45 \times 10^{-2} \text{ \AA} = 45 \times 10^{-12} \text{ m}$.
$E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{45 \times 10^{-12}} \text{ J}$.
To convert energy into $eV$, divide by $1.6 \times 10^{-19} \text{ C}$.
$E = \frac{19.86 \times 10^{-26}}{45 \times 10^{-12} \times 1.6 \times 10^{-19}} \text{ eV}$.
$E = \frac{19.86 \times 10^{-26}}{72 \times 10^{-31}} \text{ eV}$.
$E \approx 0.2758 \times 10^5 \text{ eV} = 27,580 \text{ eV}$.
Rounding to the nearest option, the maximum energy is $27,500 \text{ eV}$.

(C) The reduction reaction at the cathode is: $2H^{+} + 2e^{-} \rightarrow H_2$.
Given volume of $H_2$ gas $= 1120 \ mL = 1.12 \ L$.
Number of moles of $H_2$ gas $= \frac{1.12 \ L}{22.4 \ L/mol} = 0.05 \ mol$.
According to the reaction,$1 \ mol$ of $H_2$ requires $2 \ mol$ of electrons.
Therefore,$0.05 \ mol$ of $H_2$ requires $0.05 \times 2 = 0.1 \ mol$ of electrons.
Total charge $Q = n \times F = 0.1 \times 96500 \ C = 9650 \ C$.
Using the formula $Q = I \times t$,where $t = 1930 \ s$:
$I = \frac{Q}{t} = \frac{9650 \ C}{1930 \ s} = 5.0 \ A$.

(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the masses of substances deposited are proportional to their equivalent weights.
$\frac{\text{Mass of silver}}{\text{Mass of hydrogen}} = \frac{\text{Eq. wt. of silver}}{\text{Eq. wt. of hydrogen}}$
Given:
Mass of $H_2 = 5.04 \times 10^{-2} \ g$
Eq. wt. of $H_2 = 1.008$
Eq. wt. of $Ag = 108$
Let the mass of silver deposited be $w$.
$\frac{w}{5.04 \times 10^{-2}} = \frac{108}{1.008}$
$w = \frac{108 \times 5.04 \times 10^{-2}}{1.008}$
$w = 100 \times 5.04 \times 10^{-2} = 5.4 \ g$
Thus,the mass of silver deposited is $5.4 \ g$.

(B) The Compton shift formula is given by $\Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos \phi)$.
Given $\lambda = 0.140 \ nm = 0.140 \times 10^{-9} \ m$ and $\phi = 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the shift is $\Delta \lambda = \frac{h}{m_e c}$.
The Compton wavelength $\frac{h}{m_e c} \approx 2.426 \times 10^{-12} \ m = 0.002426 \ nm$.
Therefore,$\lambda' = \lambda + \Delta \lambda = 0.140 \ nm + 0.002426 \ nm = 0.142426 \ nm$.
Rounding to three decimal places,we get $\lambda' \approx 0.142 \ nm$.

(B) Given total charge $Q = 1 \mu C = 10^{-6} \ C$.
The charges are divided in the ratio $2:3$.
Therefore,$q_1 = \frac{2}{2+3} \times 10^{-6} \ C = 0.4 \times 10^{-6} \ C$ and $q_2 = \frac{3}{2+3} \times 10^{-6} \ C = 0.6 \times 10^{-6} \ C$.
The distance between them is $r = 1 \ m$.
The electrostatic force $F$ is given by Coulomb's Law: $F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}$.
Substituting the values: $F = (9 \times 10^9) \times \frac{(0.4 \times 10^{-6}) \times (0.6 \times 10^{-6})}{1^2}$.
$F = 9 \times 10^9 \times 0.24 \times 10^{-12} = 2.16 \times 10^{-3} \ N = 0.00216 \ N$.

(A) In the stratosphere,the following reactions occur which are responsible for the depletion of the ozone layer:
$NO + O_3 \longrightarrow NO_2 + O_2$
$CF_2Cl_2 \stackrel{h\nu}{\longrightarrow} \dot{C}F_2Cl + \dot{Cl}$
$CFCl_3 \stackrel{h\nu}{\longrightarrow} \dot{C}FCl_2 + \dot{Cl}$
$\dot{Cl} + O_3 \longrightarrow Cl\dot{O} + O_2$
$Cl\dot{O} + O \longrightarrow \dot{Cl} + O_2$
Hence,methane $(CH_4)$ is not responsible for ozone layer depletion.

(D) The given compound is an ether with the structure $C_2H_5-O-CH(CH_3)_2$.
According to $IUPAC$ nomenclature rules for ethers,they are named as alkoxyalkanes.
The larger alkyl group is taken as the parent alkane,and the smaller alkyl group along with the oxygen atom is named as the alkoxy group.
Here,the ethoxy group $(-OC_2H_5)$ is attached to the second carbon of the propane chain.
Therefore,the correct $IUPAC$ name is $2-$ethoxypropane.

(A) According to $Cahn-Ingold-Prelog$ sequence rules,the priority of groups is decided by the atomic number of the atoms directly attached to the chiral center.
$1$. The atom with the highest atomic number gets the highest priority. Here,$O$ (atomic number $8$) has higher priority than $C$ (atomic number $6$). Thus,$-OH$ is first.
$2$. For the remaining carbon-containing groups,we look at the atoms attached to the first carbon: $-COOH$ ($C$ bonded to $O, O, O$),$-CHO$ ($C$ bonded to $O, O, H$),and $-CH_2OH$ ($C$ bonded to $O, H, H$).
$3$. Comparing these,the priority order is $-COOH > -CHO > -CH_2OH$.
$4$. Therefore,the overall order is $-OH > -COOH > -CHO > -CH_2OH$.

(D) According to the Cahn-Ingold-Prelog $(CIP)$ priority rules,if the groups with higher priority are on the same side of the double bond,the configuration is '$Z$' (zusammen). If they are on opposite sides,it is '$E$' (entgegen).
For $(i)$: Priority at left carbon: $Cl > H$; Priority at right carbon: $Br > F$. Since $Cl$ and $Br$ are on the same side,it is $(Z)$.
For $(ii)$: Priority at left carbon: $Cl > H$; Priority at right carbon: $Br > F$. Since $Cl$ and $Br$ are on opposite sides,it is $(E)$.
For $(iii)$: Priority at left carbon: $Br > Cl$; Priority at right carbon: $CH_3 > H$. Since $Br$ and $CH_3$ are on the same side,it is $(Z)$.
Therefore,compounds $(i)$ and $(iii)$ have $(Z)$ configuration.

(A) The reduction of $2$-butyne with $Na / NH_3$ (liq.) (Birch reduction) proceeds via anti-addition to yield the $E$-product (trans$-2-$butene).
The reduction of $2$-butyne with $Pd / BaSO_4 + H_2$ (Lindlar's catalyst) proceeds via syn-addition to yield the $Z$-product (cis$-2-$butene).
According to the reaction scheme:
$2$-butyne $\xrightarrow{X} E$-product implies $X = Na / NH_3$ (liq.).
$2$-butyne $\xrightarrow{Y} Z$-product implies $Y = Pd / BaSO_4 + H_2$.
Therefore,$X$ and $Y$ are $Na / NH_3$ (liq.) and $Pd / BaSO_4 + H_2$ respectively.

(D) The oxidising property of $H_2O_2$ is demonstrated when it acts as an oxidising agent,meaning it gets reduced itself while oxidising another substance.
In option $A$,$B$,and $C$,$H_2O_2$ acts as a reducing agent,as it is oxidised to $O_2$.
In option $D$,$2KI + H_2SO_4 + H_2O_2 \longrightarrow K_2SO_4 + I_2 + 2H_2O$,the oxidation state of iodine increases from $-1$ in $KI$ to $0$ in $I_2$,while the oxygen in $H_2O_2$ is reduced from $-1$ to $-2$. Thus,$H_2O_2$ acts as an oxidising agent.

(C) Barium hydroxide is a strong base and dissociates as follows: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^-$.
Initially,the volume is $50 \ mL$ and the concentration is $1 \times 10^{-3} \ M$.
Upon adding $50 \ mL$ of $H_2O$,the total volume becomes $100 \ mL$.
Since the volume is doubled,the concentration of $Ba(OH)_2$ is halved: $M_2 = \frac{M_1 V_1}{V_2} = \frac{1 \times 10^{-3} \ M \times 50 \ mL}{100 \ mL} = 0.5 \times 10^{-3} \ M$.
Since each mole of $Ba(OH)_2$ produces $2$ moles of $OH^-$,the concentration of $OH^-$ is: $[OH^-] = 2 \times 0.5 \times 10^{-3} \ M = 1 \times 10^{-3} \ M$.
$pOH = -\log[OH^-] = -\log(1 \times 10^{-3}) = 3$.
$pH = 14 - pOH = 14 - 3 = 11.0$.

(A) $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In aqueous solution,it ionizes as: $CH_3COONa \rightleftharpoons CH_3COO^{-} + Na^{+}$.
The acetate ion $(CH_3COO^{-})$ undergoes anionic hydrolysis: $CH_3COO^{-} + H_2O \rightleftharpoons CH_3COOH + OH^{-}$.
Due to the production of $OH^{-}$ ions,the solution becomes alkaline (basic).
Therefore,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 N i}{2r}$.
Since the two coils carry currents in opposite directions,the net magnetic field at the center is the difference between the two fields.
Given: $N_1 = N_2 = 10$,$r_1 = 0.2 \ m$,$r_2 = 0.4 \ m$,$i_1 = 0.2 \ A$,$i_2 = 0.3 \ A$.
$B_{\text{net}} = |B_1 - B_2| = \left| \frac{\mu_0 N_1 i_1}{2r_1} - \frac{\mu_0 N_2 i_2}{2r_2} \right|$
$B_{\text{net}} = \frac{\mu_0 \times 10}{2} \left| \frac{0.2}{0.2} - \frac{0.3}{0.4} \right|$
$B_{\text{net}} = 5 \mu_0 \left| 1 - 0.75 \right|$
$B_{\text{net}} = 5 \mu_0 \times 0.25 = 5 \mu_0 \times \frac{1}{4} = \frac{5}{4} \mu_0$.

(A) The initial magnetic moment is $M = m \cdot l$,where $m$ is the pole strength.
When the wire of length $l$ is bent into a semicircle,the length $l$ becomes the arc length of the semicircle.
So,$l = \pi r$,which gives $r = \frac{l}{\pi}$.
The new distance between the poles (the diameter of the semicircle) is $l' = 2r = \frac{2l}{\pi}$.
The new magnetic moment $M'$ is given by the product of the pole strength and the new distance between the poles:
$M' = m \cdot l' = m \cdot \left(\frac{2l}{\pi}\right) = \frac{2}{\pi} (m \cdot l)$.
Since $M = m \cdot l$,we have $M' = \frac{2M}{\pi}$.

(C) The time period of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$.
When a bar magnet is cut parallel to its length into $4$ equal pieces,the mass of each piece becomes $m' = \frac{m}{4}$.
The magnetic moment of each piece becomes $M' = \frac{M}{4}$.
The moment of inertia of each piece about the axis of rotation becomes $I' = \frac{I}{4}$.
Substituting these into the formula for the new time period $T'$:
$T' = 2 \pi \sqrt{\frac{I'}{M'B}} = 2 \pi \sqrt{\frac{I/4}{(M/4)B}} = 2 \pi \sqrt{\frac{I}{MB}}$.
Thus,$T' = T = 4 ~s$.

(D) We have $\frac{x^2+x+1}{x^2+2x+1} = \frac{(x^2+2x+1) - x}{x^2+2x+1} = 1 - \frac{x}{(x+1)^2}$.
Now,express $\frac{x}{(x+1)^2}$ in partial fractions:
$\frac{x}{(x+1)^2} = \frac{A'}{x+1} + \frac{B'}{(x+1)^2} = \frac{A'(x+1) + B'}{(x+1)^2}$.
Comparing numerators,$x = A'x + (A'+B')$.
Thus,$A' = 1$ and $A'+B' = 0$,which gives $B' = -1$.
So,$\frac{x^2+x+1}{x^2+2x+1} = 1 - (\frac{1}{x+1} - \frac{1}{(x+1)^2}) = 1 - \frac{1}{x+1} + \frac{1}{(x+1)^2}$.
Comparing this with $A + \frac{B}{x+1} + \frac{C}{(x+1)^2}$,we get $A=1$,$B=-1$,and $C=1$.
Therefore,$A-B = 1 - (-1) = 2$.
Since $C=1$,$2 = 2C$.
Hence,$A-B = 2C$.

(D) Given that,$\alpha+\beta=-2$ and $\alpha^3+\beta^3=-56$.
We know that $\alpha^3+\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta)$.
Substituting the values,we get $(-2)(\alpha^2+\beta^2-\alpha\beta) = -56$,which implies $\alpha^2+\beta^2-\alpha\beta = 28$.
Also,$(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta = (-2)^2 = 4$.
Subtracting the two equations: $(\alpha^2+\beta^2+2\alpha\beta) - (\alpha^2+\beta^2-\alpha\beta) = 4 - 28$.
$3\alpha\beta = -24$,so $\alpha\beta = -8$.
The quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha+\beta)x + (\alpha\beta) = 0$.
Substituting the values,$x^2 - (-2)x + (-8) = 0$,which simplifies to $x^2+2x-8=0$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
First,simplify the powers of $\omega$:
$\omega^{10} = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$
$\omega^{23} = (\omega^3)^7 \cdot \omega^2 = 1^7 \cdot \omega^2 = \omega^2$
Now,substitute these into the expression:
$\sin \left\{\left(\omega^{10} + \omega^{23}\right) \pi - \frac{\pi}{4}\right\} = \sin \left\{\left(\omega + \omega^2\right) \pi - \frac{\pi}{4}\right\}$
Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.
Substituting this value:
$= \sin \left\{-\pi - \frac{\pi}{4}\right\} = \sin \left\{-\left(\pi + \frac{\pi}{4}\right)\right\}$
Using the property $\sin(-\theta) = -\sin(\theta)$:
$= -\sin \left(\pi + \frac{\pi}{4}\right)$
Using the property $\sin(\pi + \theta) = -\sin(\theta)$:
$= -(-\sin \frac{\pi}{4}) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

(B) Given that,${}^n P_r = 30240$ and ${}^n C_r = 252$.
We know that ${}^n P_r = {}^n C_r \times r!$.
Substituting the values,we get $30240 = 252 \times r!$.
$r! = \frac{30240}{252} = 120$.
Since $120 = 5!$,we have $r = 5$.
Now,${}^n P_5 = 30240$.
$n(n-1)(n-2)(n-3)(n-4) = 30240$.
Testing values,for $n = 10$: $10 \times 9 \times 8 \times 7 \times 6 = 30240$.
Thus,$n = 10$.
Hence,the required ordered pair is $(10, 5)$.

(C) Let the $3$ small boxes be $S_1, S_2, S_3$ and the other $6$ boxes be $B_1, B_2, B_3, B_4, B_5, B_6$.
There are $5$ balls that cannot fit into the $3$ small boxes.
These $5$ balls must be placed in the $6$ larger boxes.
The number of ways to arrange these $5$ balls in $6$ boxes is given by $^6P_5 = \frac{6!}{(6-5)!} = 6 \times 5 \times 4 \times 3 \times 2 = 720$.
After placing these $5$ balls,there are $4$ balls remaining and $4$ boxes remaining (the $3$ small boxes and $1$ unused large box).
The number of ways to arrange these $4$ balls in the remaining $4$ boxes is $4! = 24$.
Total number of arrangements = $^6P_5 \times 4! = 720 \times 24 = 17280$.

(C) The Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta l/l}$,where $F$ is the force,$A$ is the area,and $\Delta l/l$ is the strain.
Given $F = 33000 ~N$,$A = 10^{-3} ~m^2$,and $Y = 3 \times 10^{11} ~N/m^2$.
Substituting the values: $3 \times 10^{11} = \frac{33000 / 10^{-3}}{\Delta l/l}$.
Thus,the strain $\frac{\Delta l}{l} = \frac{33000}{10^{-3} \times 3 \times 10^{11}} = \frac{33 \times 10^3}{3 \times 10^8} = 11 \times 10^{-5}$.
Thermal expansion is given by $\frac{\Delta l}{l} = \alpha \Delta T$,where $\alpha = 1.1 \times 10^{-5} /{ }^{\circ} C$.
Equating the two expressions for strain: $11 \times 10^{-5} = 1.1 \times 10^{-5} \times \Delta T$.
Solving for $\Delta T$: $\Delta T = \frac{11 \times 10^{-5}}{1.1 \times 10^{-5}} = 10^{\circ} C$.

(A) The Young's modulus $Y$ is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l}$.
Rearranging for the fractional change in length $\frac{\Delta l}{l}$,we get: $\frac{\Delta l}{l} = \frac{F}{AY}$.
At the lowest position,the tension $T$ in the wire is equal to the weight of the load,$T = mg$ (assuming the velocity is negligible or the question implies static equilibrium at the lowest point for the fractional change calculation).
Given: $m = 1 ~kg$,$g = 10 ~m/s^2$,$A = 3 ~mm^2 = 3 \times 10^{-6} ~m^2$,and $Y = 10^{11} ~N/m^2$.
Substituting the values: $\frac{\Delta l}{l} = \frac{1 \times 10}{3 \times 10^{-6} \times 10^{11}} = \frac{10}{3 \times 10^5} = \frac{1}{3} \times 10^{-4} \approx 0.33 \times 10^{-4}$.
Rounding to the provided options,the correct value is $0.3 \times 10^{-4}$.

(B) Let the initial velocity be $u$. At maximum height $h$,the final velocity is $0$.
Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$.
The maximum height is $h = \frac{u^2}{2g} = \frac{(gt)^2}{2g} = \frac{gt^2}{2}$.
The body reaches the maximum height in time $t$.
While returning,the body starts from rest at height $h$ and falls to height $h/2$. The distance covered is $s = h - h/2 = h/2$.
Using $s = \frac{1}{2}gt'^2$,where $t'$ is the time taken to fall from maximum height to $h/2$:
$\frac{h}{2} = \frac{1}{2}gt'^2 \implies t'^2 = \frac{h}{g} = \frac{gt^2/2}{g} = \frac{t^2}{2}$.
Thus,$t' = \frac{t}{\sqrt{2}}$.
The total time from projection is $T = t + t' = t + \frac{t}{\sqrt{2}} = \left(1 + \frac{1}{\sqrt{2}}\right) t$.

(C) In projectile motion,the velocity vector is always tangent to the trajectory path.
At the maximum height,the vertical component of velocity $(v_y)$ becomes zero,while the horizontal component $(v_x = u \cos \theta)$ remains constant.
Since the velocity vector only has a horizontal component at the maximum height,it makes an angle of $0^\circ$ with the horizontal.
Therefore,option $C$ is correct.

(C) The nuclear force is the strong force that binds nucleons (protons and neutrons) together in the nucleus.
Experimental evidence shows that the nuclear force is charge-independent.
This means that the force between two protons,two neutrons,or a proton and a neutron is essentially the same,provided the distance between them and their spin states are identical.
Therefore,the relation is $F_{pp} = F_{nn} = F_{np}$.

(D) Simple harmonic motion is a periodic motion where the particle repeats its position and velocity after every time interval equal to the time period $T$.
Given that the particle starts from an initial position,after one time period $T$,the particle returns to its initial position,meaning the displacement is $0$.
After $2T$ (which is an integer multiple of $T$),the particle will have completed two full cycles.
Therefore,the particle will return to its initial position again,and the net displacement from the initial position will be $0$.

(C) According to the Lewis concept,a substance that can accept a lone pair of electrons is called a Lewis acid.
Boron halides (e.g.,$BX_3$) have only $6$ electrons in the valence shell of the boron atom.
Due to this electron deficiency,they can accept a lone pair of electrons from a donor to complete their octet,thus acting as Lewis acids.

(D) Orthosilicic acid $(H_4SiO_4)$,on heating at high temperature,loses two water molecules to form silica $(SiO_2)$ as $A$.
$H_4SiO_4 \xrightarrow[1000^{\circ}C]{-2H_2O} SiO_2 (A)$
Silica $(SiO_2)$ on reduction with carbon at high temperature gives carborundum $(SiC)$ as $B$ and releases carbon monoxide $(CO)$.
$SiO_2 + 3C \xrightarrow{\Delta} SiC (B) + 2CO$
Therefore,$B$ is carborundum.

(B) The structure of peroxodisulphuric acid $(H_2S_2O_8)$ is $HO-SO_2-O-O-SO_2-OH$.
In this structure,there are $4$ $S=O$ bonds (each containing $1$ $\sigma$ and $1$ $\pi$ bond),$2$ $S-OH$ bonds,$2$ $S-O$ bonds,and $1$ $O-O$ bond.
Total $\sigma$ bonds = $4$ (from $S=O$) + $2$ (from $S-OH$) + $2$ (from $S-O$) + $1$ (from $O-O$) + $2$ (from $O-H$) = $11$ $\sigma$ bonds.
Total $\pi$ bonds = $4$ (from $S=O$) = $4$ $\pi$ bonds.
Hence,it contains $11$ $\sigma$ and $4$ $\pi$ bonds.

(D) An oxidising agent is a substance that undergoes reduction (gain of electrons or decrease in oxidation state) and causes the oxidation of another substance (loss of electrons,addition of oxygen,or removal of hydrogen).
In all the given reactions,the oxidation state of chlorine decreases from $0$ (in $Cl_2$) to $-1$ (in $HCl$),which means chlorine is reduced.
$(i)$ $CH_3CH_2OH$ loses hydrogen to form $CH_3CHO$,so $Cl_2$ acts as an oxidising agent.
$(ii)$ $CH_3CHO$ loses hydrogen to form $CCl_3CHO$,so $Cl_2$ acts as an oxidising agent.
$(iii)$ $CH_4$ loses hydrogen to form $CH_3Cl$,so $Cl_2$ acts as an oxidising agent.
Therefore,in all these reactions,chlorine acts as an oxidising agent.

(B) The oxidizing power of halogens decreases down the group as their reduction potential decreases. $F_2$ is the strongest oxidizing agent,followed by $Cl_2$,$Br_2$,and $I_2$.
$A$ halogen with a higher reduction potential can displace a halogen with a lower reduction potential from its salt solution.
Since $F_2$ has the highest reduction potential,it can oxidize $Cl^-$,$Br^-$,and $I^-$.
However,$Cl_2$ cannot oxidize $F^-$ because $F_2$ is a stronger oxidizing agent than $Cl_2$.
Therefore,the reaction $Cl_2 + 2F^{-} \longrightarrow 2Cl^{-} + F_2$ does not occur.

(C) Let $E = \sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.
$E = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}}$
$E = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$
Multiply and divide by $2$:
$E = 2 \left( \frac{\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
$E = 2 \left( \frac{\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:
$E = 2 \left( \frac{\sin(60^{\circ} - 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}} \right) = 2 \left( \frac{\sin 40^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}} \right)$
Multiply numerator and denominator by $2$:
$E = 4 \left( \frac{\sin 40^{\circ}}{2 \sin 20^{\circ} \cos 20^{\circ}} \right) = 4 \left( \frac{\sin 40^{\circ}}{\sin 40^{\circ}} \right) = 4$.

(B) Given that $\alpha+\beta+\gamma=2 \theta$,so $\theta = \frac{\alpha+\beta+\gamma}{2}$.
Let $S = \cos \theta + \cos (\theta-\alpha) + \cos (\theta-\beta) + \cos (\theta-\gamma)$.
Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$:
$S = [\cos \theta + \cos (\theta-\gamma)] + [\cos (\theta-\alpha) + \cos (\theta-\beta)]$
$S = 2 \cos \frac{2\theta-\gamma}{2} \cos \frac{\gamma}{2} + 2 \cos \frac{2\theta-\alpha-\beta}{2} \cos \frac{\beta-\alpha}{2}$
Since $2\theta = \alpha+\beta+\gamma$,we have $2\theta-\gamma = \alpha+\beta$ and $2\theta-\alpha-\beta = \gamma$.
$S = 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\gamma}{2} + 2 \cos \frac{\gamma}{2} \cos \frac{\beta-\alpha}{2}$
$S = 2 \cos \frac{\gamma}{2} [\cos \frac{\alpha+\beta}{2} + \cos \frac{\beta-\alpha}{2}]$
Using $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ again:
$S = 2 \cos \frac{\gamma}{2} [2 \cos \frac{\beta}{2} \cos \frac{\alpha}{2}]$
$S = 4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}$.

(B) Given that,$A=35^{\circ}, B=15^{\circ}$ and $C=40^{\circ}$.
Since $A+B+C = 35^{\circ}+15^{\circ}+40^{\circ} = 90^{\circ}$,we use the identity for $\tan(A+B+C)$:
$\tan(A+B+C) = \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A)}$.
As $A+B+C = 90^{\circ}$,$\tan(90^{\circ})$ is undefined,which implies the denominator must be zero:
$1 - (\tan A \tan B + \tan B \tan C + \tan C \tan A) = 0$.
Therefore,$\tan A \tan B + \tan B \tan C + \tan C \tan A = 1$.

(B) Given equation is $\cos 2x + 2 \cos^2 x = 2$.
Using the identity $\cos 2x = 2 \cos^2 x - 1$,we get:
$(2 \cos^2 x - 1) + 2 \cos^2 x = 2$
$4 \cos^2 x - 1 = 2$
$4 \cos^2 x = 3$
$\cos^2 x = \frac{3}{4}$
$\cos x = \pm \frac{\sqrt{3}}{2}$
For $\cos x = \frac{\sqrt{3}}{2}$,$x = 2n\pi \pm \frac{\pi}{6}$.
For $\cos x = -\frac{\sqrt{3}}{2}$,$x = 2n\pi \pm \frac{5\pi}{6}$.
Combining these,the general solution is $x = n\pi \pm \frac{\pi}{6}$ for $n \in Z$.

(A) The correct matching is:
$A$. $Nylon$ $6,6$ is a polyamide formed from $Adipic$ $acid$ and $Hexamethylenediamine$ $(v)$.
$B$. $Terylene$ (also known as $Dacron$) is a polyester formed from $Ethylene$ $glycol$ and $Terephthalic$ $acid$ $(i)$.
$C$. $Buna-S$ is a copolymer of $1,3-Butadiene$ and $Styrene$ $(ii)$.
$D$. $Neoprene$ is a polymer of $Chloroprene$ $(iii)$.
Therefore,the correct sequence for $A, B, C, D$ is $(v, i, ii, iii)$.

(A) The given compounds are:
$1. [Co(NH_3)_5SO_4]Br \rightleftharpoons [Co(NH_3)_5SO_4]^+ + Br^-$
$2. [Co(NH_3)_5Br]SO_4 \rightleftharpoons [Co(NH_3)_5Br]^{2+} + SO_4^{2-}$
Since both compounds have the same molecular formula but produce different ions in an aqueous solution,they exhibit ionisation isomerism.

(A) The reducing character of the hydrides of group $V$ elements depends upon the stability of the $E-H$ bond.
As we move down the group,the atomic size of the central element increases,which leads to a decrease in the bond dissociation enthalpy and stability of the hydrides.
Consequently,the ability to release hydrogen atoms increases,making the reducing character increase down the group.
Thus,the correct order of reducing abilities is $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.

(B) Given,mass defect $\Delta m = 0.081 \ u$.
Total number of nucleons $A = 11$.
Binding energy $BE = \Delta m \times 931 \ MeV/u = 0.081 \times 931 = 75.411 \ MeV$.
Average binding energy per nucleon $= \frac{BE}{A} = \frac{75.411}{11} = 6.855 \ MeV \approx 6.85 \ MeV$.

(C) Acetic acid on reduction with lithium aluminium hydride $(LiAlH_4)$ gives ethyl alcohol.
$CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH$
Acetic acid on reduction with $HI$ and red $P$ gives ethane.
$CH_3COOH \xrightarrow{\text{Red } P + HI} CH_3-CH_3$
Therefore,reagent $A$ is $LiAlH_4$ and reagent $B$ is $HI + \text{red } P$.

(D) The reaction of hydrogen halides $(HX)$ with ethyl alcohol $(C_2H_5OH)$ involves the cleavage of the $C-O$ bond.
The reactivity depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond dissociation energy of the $H-X$ bond decreases,making the bond easier to break.
Therefore,the reactivity order is $HI > HBr > HCl > HF$.

(B) Acetone $(CH_3COCH_3)$ reacts with methyl magnesium bromide $(CH_3MgBr)$ to form an addition complex.
This complex is $(CH_3)_3COMgBr$.
Upon acid hydrolysis (decomposition with acid),the complex undergoes protonation to yield a tertiary alcohol,$2$-methylpropan-$2$-ol,which is $(CH_3)_3COH$,and $Mg(OH)Br$.
Thus,$X$ is $(CH_3)_3COH$.

(C) The reduction of nitrobenzene with lithium aluminium hydride $(LiAlH_4)$ in an alkaline medium or specific conditions leads to the formation of azobenzene as the major product.
The reaction is represented as:
$2 C_6H_5NO_2 \xrightarrow{LiAlH_4} C_6H_5-N=N-C_6H_5$ (Azobenzene).

(A) Sucrose undergoes hydrolysis in the presence of dilute aqueous sulphuric acid to produce an equimolar mixture of $D-(+)$-glucose and $D-(-)$-fructose.
The chemical reaction is:
$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H_2SO_4} C_6H_{12}O_6 + C_6H_{12}O_6$
($D-(+)$-glucose) ($D-(-)$-fructose)
The molar ratio of the products is $1:1$.
Sucrose is dextrorotatory,but the resulting mixture is laevorotatory because the specific rotation of $D-(-)$-fructose $(-92.4^{\circ})$ is greater in magnitude than that of $D-(+)$-glucose $(+52.7^{\circ})$.

(C) From the given energy profile diagram:
$E_a$ = activation energy of forward reaction
$E_a^{\prime}$ = activation energy of backward reaction
$E_t$ = threshold energy
$1$. Since the energy level of product $B$ is higher than reactant $A$,the reaction is endothermic.
$2$. The activation energy of the forward reaction $(E_a)$ is the difference between threshold energy $(E_t)$ and energy of reactant $(E_R)$.
$3$. The activation energy of the backward reaction $(E_a^{\prime})$ is the difference between threshold energy $(E_t)$ and energy of product $(E_p)$.
$4$. From the diagram,$E_a > E_a^{\prime}$.
$5$. The relationship is $E_a = E_a^{\prime} + \Delta E$,where $\Delta E$ is the heat of reaction.
$6$. The threshold energy $(E_t)$ is always greater than or equal to the activation energy ($E_a$ or $E_a^{\prime}$),as it represents the minimum energy required for the reaction to occur. Therefore,the statement that 'threshold energy is less than that of activation energy' is incorrect.

(A) Based on the classification of amines:
$(A)$ $CH_3CH_2CH_2CH_2NH_2$ contains the $-NH_2$ group attached to one alkyl group,making it a primary amine $(1^\circ)$. Thus,$(A) \rightarrow (v)$.
$(B)$ $(CH_3CH_2)_2NH$ contains the $-NH-$ group attached to two alkyl groups,making it a secondary amine $(2^\circ)$. Thus,$(B) \rightarrow (i)$.
$(C)$ $(CH_3)_3N$ has the nitrogen atom attached to three alkyl groups,making it a tertiary amine $(3^\circ)$. Thus,$(C) \rightarrow (ii)$.
$(D)$ $C_6H_5NH_2$ (Aniline) is an amine where the nitrogen is directly attached to a benzene ring,making it an aromatic amine. Thus,$(D) \rightarrow (iii)$.
Therefore,the correct sequence is $A-(v), B-(i), C-(ii), D-(iii)$.

(A) The given compounds are:
$1. [Co(NH_3)_5SO_4]Br \rightleftharpoons [Co(NH_3)_5SO_4]^+ + Br^-$
$2. [Co(NH_3)_5Br]SO_4 \rightleftharpoons [Co(NH_3)_5Br]^{2+} + SO_4^{2-}$
Since both compounds have the same molecular formula but produce different ions in an aqueous solution,they exhibit ionisation isomerism.

(C) The reduction reaction at the cathode is: $2H^+ + 2e^- \rightarrow H_2$.
Number of moles of $H_2$ gas collected = $\frac{1120 \,mL}{22400 \,mL/mol} = 0.05 \,mol$.
From the stoichiometry of the reaction, $1 \,mol$ of $H_2$ requires $2 \,mol$ of electrons.
Therefore, $0.05 \,mol$ of $H_2$ requires $0.05 \times 2 = 0.1 \,mol$ of electrons.
Total charge $Q = n \times F = 0.1 \,mol \times 96500 \,C/mol = 9650 \,C$.
Using the relation $Q = I \times t$, where $t = 1930 \,s$:
$I = \frac{Q}{t} = \frac{9650 \,C}{1930 \,s} = 5.0 \,A$.

(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes in series,the masses of substances deposited are proportional to their equivalent weights.
$\frac{\text{Mass of Ag}}{\text{Mass of } H_2} = \frac{\text{Eq. wt. of Ag}}{\text{Eq. wt. of } H_2}$
Given:
Mass of $H_2 = 5.04 \times 10^{-2} \ g$
Eq. wt. of $H_2 = 1.008$
Eq. wt. of $Ag = 108$
Let the mass of silver deposited be $w$.
$\frac{w}{5.04 \times 10^{-2}} = \frac{108}{1.008}$
$w = \frac{108 \times 5.04 \times 10^{-2}}{1.008}$
$w = 107.14 \times 5.04 \times 10^{-2} \approx 5.4 \ g$
Therefore,the mass of silver deposited is $5.4 \ g$.

(A) The reducing character of the hydrides of group $15$ elements depends upon the thermal stability of the $E-H$ bond.
As we move down the group,the bond dissociation enthalpy decreases due to an increase in the size of the central atom.
Consequently,the stability of the hydrides decreases,and the ease of releasing hydrogen atoms increases.
Therefore,the reducing character increases down the group.
The correct order is: $NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3$.

(A) Given: Angle of diffraction $(2 \theta) = 90^{\circ}$.
Therefore,$\theta = 45^{\circ}$.
Distance between two planes,$d = 2.28 \ \text{Å}$.
Order of diffraction,$n = 2$.
Bragg's equation is given by $n \lambda = 2 d \sin \theta$.
Substituting the values: $2 \times \lambda = 2 \times 2.28 \times \sin 45^{\circ}$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.7071$,we have $2 \lambda = 2 \times 2.28 \times 0.7071$.
$\lambda = 2.28 \times 0.7071 = 1.612 \ \text{Å}$.

(C) Given:
Weight of non-volatile solute,$w = 25 \ g$
Weight of solvent,$W = 100 \ g$
Lowering of vapour pressure,$p^{\circ} - p_s = 0.225 \ mm$
Vapour pressure of pure solvent,$p^{\circ} = 17.5 \ mm$
Molecular weight of solvent $(H_2O)$,$M = 18 \ g/mol$
Molecular weight of solute,$m = ?$
According to Raoult's law for non-volatile solutes:
$\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{w \times M}{m \times W}$
Substituting the values:
$\frac{0.225}{17.5} = \frac{25 \times 18}{m \times 100}$
$m = \frac{25 \times 18 \times 17.5}{0.225 \times 100}$
$m = \frac{25 \times 18 \times 17.5}{22.5}$
$m = 350 \ g/mol$

(B) The ratio of weight average molecular weight to the number average molecular weight is defined as the poly dispersity index $(PDI)$.
$PDI = \frac{\bar{M}_w}{\bar{M}_n}$
where,
$\bar{M}_w = \text{weight average molecular weight}$
$\bar{M}_n = \text{number average molecular weight}$
For natural monodispersed polymers,$PDI = 1$,whereas for synthetic polymers,$PDI > 1$.

(B) Given,mass defect $\Delta m = 0.081 \ u$.
Number of nucleons $A = 11$.
Binding energy $= 931 \times \Delta m \ MeV = 931 \times 0.081 \ MeV = 75.411 \ MeV$.
Average binding energy $= \frac{\text{Binding energy}}{A} = \frac{75.411}{11} \ MeV = 6.85 \ MeV$.

(C) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = k p^{1/n}$.
Taking the logarithm on both sides,we get: $\log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log p$.
This equation is of the form $y = mx + c$,where $y = \log (x / m)$,$x = \log p$,slope $m = 1/n$,and intercept $c = \log k$.
Therefore,plotting $\log (x / m)$ on the $y$-axis and $\log p$ on the $x$-axis yields a straight line with a positive slope of $1/n$ and an intercept of $\log k$ on the $y$-axis.