WBJEE 2021 Chemistry Question Paper with Answer and Solution

(D) In structures $III$ and $IV$,the nitrogen atom of the nitro group is shown to be bonded to five atoms/groups (pentavalent).
Nitrogen belongs to the second period and does not have vacant $d-$orbitals in its valence shell,so it cannot expand its covalency beyond $4$.
Therefore,structures $III$ and $IV$ are chemically impossible and cannot represent the resonating structures of $p-$nitro$-N, N-$dimethylaniline.

(A) Delocalisation of $\pi$-electrons occurs in molecules that exhibit resonance.
$O_{3}$ (ozone) exists as a resonance hybrid of two canonical structures,where the $\pi$-electrons are delocalised over the three oxygen atoms.
$CO$ and $HCN$ have localised $\pi$-bonds and do not exhibit resonance-based delocalisation of $\pi$-electrons.
Therefore,the correct option is $A$.

(B) In the $H_3O^{+}$ ion,the central oxygen atom is $sp^3$ hybridized.
It contains $3$ bond pairs and $1$ lone pair.
According to the $VSEPR$ theory,the presence of one lone pair in a tetrahedral electron geometry results in a pyramidal molecular shape.

(D) The molecular formula of $p$-nitrobenzonitrile is $C_{7}H_{4}N_{2}O_{2}$.
In this molecule,the benzene ring is planar ($sp^{2}$ hybridized carbons).
The nitro group $(-NO_{2})$ is attached to the benzene ring,and due to resonance,it tends to lie in the same plane as the benzene ring.
The cyano group $(-CN)$ is linear ($sp$ hybridized carbon) and is also in the same plane as the benzene ring.
Therefore,all $15$ atoms ($7$ carbons,$4$ hydrogens,$2$ nitrogens,and $2$ oxygens) lie in the same plane.

(A) catalyst provides an alternative reaction pathway with a lower activation energy,which increases the rate of both the forward and backward reactions equally.
Since the rate of both reactions increases by the same factor,the equilibrium position remains unchanged.
The equilibrium constant $(K_{eq})$ is a function of temperature only for a given reaction.
Since the temperature remains constant at $2000 \ K$,the equilibrium constant remains $4 \times 10^{-4}$.

(C) For two wires of length $l$ and cross-sectional area $A$,the resistance is given by $R = \frac{l}{\sigma A}$.
When connected in series,the equivalent resistance is $R_{eq} = R_{1} + R_{2}$.
Substituting the values,we get $\frac{2l}{\sigma_{eq} A} = \frac{l}{\sigma_{1} A} + \frac{l}{\sigma_{2} A}$.
Dividing both sides by $\frac{l}{A}$,we get $\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}}$.
$\frac{2}{\sigma_{eq}} = \frac{\sigma_{1} + \sigma_{2}}{\sigma_{1} \sigma_{2}}$.
Therefore,the effective conductivity is $\sigma_{eq} = \frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1} + \sigma_{2}}$.

(B) According to the Law of Equivalence,the number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent.
Number of equivalents of $Fe^{2+}$ = Number of equivalents of $MnO_{4}^{-}$ = Number of equivalents of $Cr_{2}O_{7}^{2-}$.
For $MnO_{4}^{-}$ in acidic medium: $MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$. The $n$-factor is $5$.
Number of equivalents of $MnO_{4}^{-} = x \times 5 = 5x$.
For $Cr_{2}O_{7}^{2-}$ in acidic medium: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_{2}O$. The $n$-factor is $6$.
Let the number of moles of $Cr_{2}O_{7}^{2-}$ be $y$.
Number of equivalents of $Cr_{2}O_{7}^{2-} = y \times 6 = 6y$.
Equating the equivalents: $6y = 5x$.
Therefore,$y = \frac{5}{6}x = 0.833x \approx 0.83x$.

(D) The acidity of a compound depends on the stability of its conjugate base.
$1$. Acetic acid $(CH_3COOH)$ is the most acidic because its conjugate base (acetate ion,$CH_3COO^-$) is stabilized by resonance with two equivalent oxygen atoms.
$2$. $p-$Nitrophenol is the next most acidic because the phenoxide ion is stabilized by resonance,and the $-NO_2$ group at the para position exerts a strong $-I$ and $-M$ effect,further stabilizing the negative charge.
$3$. Ethanol $(CH_3CH_2OH)$ is less acidic than phenols because the ethoxide ion $(CH_3CH_2O^-)$ is destabilized by the $+I$ effect of the ethyl group.
$4$. Acetylene $(HC \equiv CH)$ is the least acidic among these as the negative charge resides on an $sp$ hybridized carbon atom,which is less electronegative than oxygen.
Thus,the correct order of acidity is: $\text{acetic acid} > p-\text{nitrophenol} > \text{ethanol} > \text{acetylene}$.

(C) Pair $1$: Both structures represent the same molecule. By rotating the second structure by $180^{\circ}$ in the plane,it superimposes on the first. Thus,they are homomers (identical).
Pair $2$: Both structures are $m$-nitrotoluene. They are identical (homomers).
Pair $3$: The first structure is $1,1$-dichloro-$2$-methylprop-$1$-ene and the second is $1,2$-dichloro-$2$-methylprop-$1$-ene. These are constitutional isomers (specifically,positional isomers).

(C) Step $1$: $Me-C \equiv C-Me$ (but$-2-$yne) reacts with $Na/NH_3(liq.)$ to form $trans$-but$-2-$ene.
Step $2$: $trans$-but$-2-$ene reacts with dilute alkaline $KMnO_4$ (Baeyer's reagent),which performs $syn$-hydroxylation.
The $syn$-addition of two $-OH$ groups to a $trans$-alkene results in a racemic mixture of butane$-2,3-$diol.

(D) In $N_2H_4$ (hydrazine),each nitrogen atom is bonded to two hydrogen atoms and one nitrogen atom,with one lone pair. The steric number for each $N$ atom is $3 + 1 = 4$,which corresponds to $sp^3$ hybridisation.
In $H_2O_2$ (hydrogen peroxide),each oxygen atom is bonded to one hydrogen atom and one oxygen atom,with two lone pairs. The steric number for each $O$ atom is $2 + 2 = 4$,which corresponds to $sp^3$ hybridisation.
Therefore,both compounds show similarity in the hybridisation of their central atoms $(sp^3)$.

(C) At equivalence point,the moles of base equal the moles of acid: $n_{base} = n_{acid}$.
$0.4 \ M \times 2.5 \ mL = \frac{2}{15} \ M \times V_{acid}$.
$V_{acid} = \frac{0.4 \times 2.5 \times 15}{2} = 7.5 \ mL$.
Total volume $= 2.5 \ mL + 7.5 \ mL = 10 \ mL$.
Concentration of salt $(C)$ $= \frac{0.4 \times 2.5}{10} = 0.1 \ M$.
The salt formed is a salt of a weak base and a strong acid,which undergoes cationic hydrolysis.
The formula for $[H^{+}]$ is $[H^{+}] = \sqrt{\frac{K_{w} \times C}{K_{b}}}$.
$[H^{+}] = \sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}} = \sqrt{10^{-3}} = \sqrt{10 \times 10^{-4}} = 3.16 \times 10^{-2} \ M \approx 3.2 \times 10^{-2} \ M$.

(C, D) buffer solution is formed by a weak acid and its conjugate base or a weak base and its conjugate acid.
$1$. $CH_{3}COOH$ (weak acid) and $CH_{3}COONa$ (salt of weak acid and strong base) form an acidic buffer. Thus,$(D)$ is correct.
$2$. When $HNO_{3}$ (strong acid) is mixed with $CH_{3}COONa$ (salt of weak acid) in a ratio such that the salt is in excess,the reaction $HNO_{3} + CH_{3}COO^{-} \rightarrow CH_{3}COOH + NO_{3}^{-}$ occurs. This results in a mixture of the weak acid $(CH_{3}COOH)$ and its conjugate base $(CH_{3}COO^{-})$,which acts as a buffer. Thus,$(C)$ is also correct.

(D) For $MX$: $K_{sp} = S^2 = 4.0 \times 10^{-8} \Rightarrow S = 2 \times 10^{-4} \text{ mol dm}^{-3}$.
For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14}$ $\Rightarrow S^3 = 8 \times 10^{-15}$ $\Rightarrow S = 2 \times 10^{-5} \text{ mol dm}^{-3}$.
For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15}$ $\Rightarrow S^4 = 10^{-16}$ $\Rightarrow S = 1 \times 10^{-4} \text{ mol dm}^{-3}$.
Comparing the solubilities: $2 \times 10^{-4} > 1 \times 10^{-4} > 2 \times 10^{-5}$.
Thus,the order of solubility is $MX > M_3X > MX_2$.

(B) For $BaSO_4$,the solubility product constant $K_{sp} = s^2 = (4 \times 10^{-5})^2 = 1.6 \times 10^{-9}$.
In $0.1 \ M \ Na_2SO_4$,the concentration of the common ion $[SO_4^{2-}] = 0.1 \ M$.
Let the new solubility be $s'$.
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = s' \times 0.1$.
$1.6 \times 10^{-9} = s' \times 0.1$.
$s' = 1.6 \times 10^{-8} \ M$.

(A) Cyclo$[18]$ carbon $(C_{18})$ is a cyclic allotrope of carbon consisting of $18$ carbon atoms.
In this structure,the carbon atoms are connected by alternating single and triple bonds.
Since there are $18$ carbon atoms in the ring,and each triple bond involves two carbon atoms while each single bond connects two adjacent triple-bonded units,the structure consists of $9$ triple bonds and $9$ single bonds.
Therefore,the total number of triple bonds is $9$.

(C) The reaction between silver nitrate $(AgNO_{3})$ and phosphorous acid $(H_{3}PO_{3})$ is a redox reaction where silver ions are reduced to metallic silver.
The balanced chemical equation is:
$2 \ AgNO_{3} + H_{3}PO_{3} + H_{2}O \longrightarrow H_{3}PO_{4} + 2 \ Ag + 2 \ HNO_{3}$
As shown in the equation,metallic silver $(Ag)$ is produced.

(D) The process described is the Solvay process for the manufacture of sodium carbonate.
Passing $CO_{2}$ $(X)$ through an ammoniacal brine solution $(NaCl + NH_{3} + H_{2}O)$ yields sodium bicarbonate $(Y)$ as a white precipitate.
$NH_{3} + CO_{2} + H_{2}O \longrightarrow NH_{4}HCO_{3}$
$NH_{4}HCO_{3} + NaCl \longrightarrow NaHCO_{3} \downarrow (Y) + NH_{4}Cl$
Upon heating,$NaHCO_{3}$ decomposes: $2NaHCO_{3} \xrightarrow{\Delta} Na_{2}CO_{3} (Z) + CO_{2} + H_{2}O$.
The weight loss percentage for $2NaHCO_{3} (2 \times 84 = 168 \ g)$ to $Na_{2}CO_{3} (106 \ g)$ is $\frac{168 - 106}{168} \times 100 \approx 36.9 \% \approx 37 \%$.
Thus,$(X) = CO_{2}$,$(Y) = NaHCO_{3}$,and $(Z) = Na_{2}CO_{3}$.

(D) The root mean square velocity of a gas is given by $u = \sqrt{\frac{3RT}{M}}$.
Since the temperature $T$ is the same for both gases,$u \propto \frac{1}{\sqrt{m}}$.
Therefore,$\frac{u_{1}}{u_{2}} = \sqrt{\frac{m_{2}}{m_{1}}}$.
Squaring both sides,we get $\frac{u_{1}^{2}}{u_{2}^{2}} = \frac{m_{2}}{m_{1}}$.
Cross-multiplying gives $m_{1} u_{1}^{2} = m_{2} u_{2}^{2}$.

(B) The given equation of the incident ray is $x+\sqrt{3}y=\sqrt{3}$.
To find the point of incidence $A$ on the $x$-axis,set $y=0$ in the equation: $x+\sqrt{3}(0)=\sqrt{3} \Rightarrow x=\sqrt{3}$. Thus,$A \equiv (\sqrt{3}, 0)$.
To find another point $B$ on the incident ray,set $x=0$: $0+\sqrt{3}y=\sqrt{3} \Rightarrow y=1$. Thus,$B \equiv (0, 1)$.
The reflected ray passes through $A(\sqrt{3}, 0)$ and the reflection of point $B$ across the $x$-axis,which is $B' \equiv (0, -1)$.
The equation of the line passing through $A(\sqrt{3}, 0)$ and $B'(0, -1)$ is given by the two-point form: $\frac{y-0}{x-\sqrt{3}} = \frac{-1-0}{0-\sqrt{3}}$.
$\frac{y}{x-\sqrt{3}} = \frac{-1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.
$\sqrt{3}y = x-\sqrt{3}$.

(A) The de Broglie wavelength is given by $\lambda = \frac{h}{p} = \frac{h}{m \cdot v}$.
For a gas,the most probable velocity is $C_{mp} = \sqrt{\frac{2KT}{m}}$.
Thus,$\lambda = \frac{h}{m \sqrt{\frac{2KT}{m}}} = \frac{h}{\sqrt{2mKT}}$.
Given $T_{He} = -73 + 273 = 200 \ K$ and $T_{Ne} = 727 + 273 = 1000 \ K$.
$\lambda_{He} = M \times \lambda_{Ne}$ $\Rightarrow \frac{h}{\sqrt{2 \times 4 \times K \times 200}} = M \times \frac{h}{\sqrt{2 \times 20 \times K \times 1000}}$.
$\frac{1}{\sqrt{1600K}} = M \times \frac{1}{\sqrt{40000K}}$.
$\frac{1}{40\sqrt{K}} = M \times \frac{1}{200\sqrt{K}}$.
$M = \frac{200}{40} = 5$.

(C) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative.
The relationship is given by the equation $\Delta G = \Delta H - T \Delta S$.
For $\Delta G$ to be negative at all temperatures $(T)$,the enthalpy change $\Delta H$ must be negative (exothermic) and the entropy change $\Delta S$ must be positive (increase in disorder).
Thus,the condition is $\Delta H < 0$ and $\Delta S > 0$.

(C) $1$. $\text{Phenol}$ reacts with dilute $HNO_3$ to form a mixture of $o$-nitrophenol and $p$-nitrophenol. The major product is $p$-nitrophenol $(X)$.
$2$. $p$-nitrophenol $(X)$ is reduced using $Zn/HCl$ to form $p$-aminophenol.
$3$. $p$-aminophenol reacts with $1 \ equiv$ of acetic anhydride $(CH_3CO)_2O$. The amino group $(-NH_2)$ is more nucleophilic than the hydroxyl group $(-OH)$, so acetylation occurs at the nitrogen atom to form $p$-acetamidophenol $(Y)$.

(D) $1$. Reaction of benzaldehyde with methanol in the presence of dry $HCl$ gas leads to the formation of an acetal,$A$,which is benzaldehyde dimethyl acetal,$Ph-CH(OMe)_2$.
$2$. Treatment of the acetal $A$ with dilute $HCl$ results in its hydrolysis back to benzaldehyde $(Ph-CHO)$ and methanol.
$3$. The subsequent reaction of benzaldehyde with acetic anhydride $((CH_3CO)_2O)$ in the presence of sodium acetate $(CH_3COONa)$ is the Perkin condensation reaction,which yields cinnamic acid $(Ph-CH=CH-COOH)$ as the final product $B$.

(D) $1$. Formation of $X$: The reaction of $p$-bromotoluene with $Mg$ in ether forms a Grignard reagent,$p-CH_3-C_6H_4-MgBr$. This reacts with acetaldehyde $(CH_3CHO)$ followed by hydrolysis to give $1-(p-tolyl)ethanol$. The haloform reaction $(Br_2/NaOH)$ on this secondary alcohol oxidizes it to $p$-methylacetophenone,which then undergoes haloform reaction again to produce $p$-toluic acid $(X)$.
$2$. Formation of $Y$: $p$-toluic acid $(X)$ reacts with $SOCl_2$ to form $p$-toluoyl chloride,which on reaction with $NH_3$ gives $p$-toluamide. Finally,the Hoffman bromamide reaction $(Br_2/NaOH)$ converts the amide into $p$-toluidine $(Y)$.

(D) When two different amino acids,glycine $(G)$ and alanine $(A)$,react to form dipeptides,they can combine in different sequences to form different products.
The possible combinations are:
$1.$ Glycine + Glycine $\rightarrow$ Glycylglycine $(G-G)$
$2.$ Alanine + Alanine $\rightarrow$ Alanylalanine $(A-A)$
$3.$ Glycine + Alanine $\rightarrow$ Glycylalanine $(G-A)$
$4.$ Alanine + Glycine $\rightarrow$ Alanylglycine $(A-G)$
Since the question asks for dipeptides obtained from glycine and alanine,it implies all combinations involving these two amino acids (including self-condensation). Thus,all four listed dipeptides can be formed.

(A) For first-order reaction:
$t_{1/2} = \frac{0.693}{k_{1}} \Rightarrow k_{1} = \frac{0.693}{40} \ s^{-1} \quad (I)$
For zero-order reaction:
$t_{1/2} = \frac{[R]_{0}}{2k_{0}} \Rightarrow k_{0} = \frac{1.386}{2 \times 20} \ mol \ dm^{-3} \ s^{-1} \quad (II)$
Calculating the ratio $\frac{k_{1}}{k_{0}}$:
$\frac{k_{1}}{k_{0}} = \frac{0.693 / 40}{1.386 / 40} = \frac{0.693}{1.386} = 0.5 \ mol^{-1} \ dm^{3}$

(A) In $K_3[Fe(CN)_6]$,the oxidation state of $Fe$ is $+3$.
Electronic configuration of $Fe^{3+}$ is $[Ar] 3d^5$.
Since $CN^-$ is a strong field ligand,pairing occurs,resulting in $t_{2g}^5 e_g^0$ configuration with $1$ unpaired electron.
In $K_4[Fe(CN)_6]$,the oxidation state of $Fe$ is $+2$.
Electronic configuration of $Fe^{2+}$ is $[Ar] 3d^6$.
Due to the strong field ligand $CN^-$,pairing occurs,resulting in $t_{2g}^6 e_g^0$ configuration with $0$ unpaired electrons.

(D) The magnetic moment depends on the number of unpaired electrons $(n)$. The formula is $\mu = \sqrt{n(n+2)} \ BM$.
For $[Cr(H_{2}O)_{6}]^{3+}$: $Cr$ is in $+3$ oxidation state. Electronic configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$. Number of unpaired electrons $(n)$ = $3$.
Now,let us check the number of unpaired electrons for each option:
$A$) $[Cu(H_{2}O)_{6}]^{2+}$: $Cu^{2+}$ is $[Ar] 3d^{9}$,$n = 1$.
$B$) $[Mn(H_{2}O)_{6}]^{3+}$: $Mn^{3+}$ is $[Ar] 3d^{4}$,$n = 4$.
$C$) $[Fe(H_{2}O)_{6}]^{3+}$: $Fe^{3+}$ is $[Ar] 3d^{5}$,$n = 5$.
$D$) $[Mn(H_{2}O)_{6}]^{4+}$: $Mn^{4+}$ is $[Ar] 3d^{3}$,$n = 3$.
Since $[Mn(H_{2}O)_{6}]^{4+}$ has $3$ unpaired electrons,it has the same magnetic moment as $[Cr(H_{2}O)_{6}]^{3+}$.

(C) When a concentrated solution of $Co(NO_3)_2$ reacts with $NaNO_2$ in $50\%$ acetic acid,the hexanitritocobaltate$(III)$ complex ion,$[Co(NO_2)_6]^{3-}$,is formed.
Upon adding a salt containing the potassium ion $(K^+)$,a yellow precipitate of potassium hexanitritocobaltate$(III)$,$K_3[Co(NO_2)_6]$,is produced.
This reaction is a standard qualitative test for the identification of potassium ions.
Therefore,metal $M$ is potassium $(K)$.

(C) The reduction reaction for a hydrogen half-cell is: $2H^{+} + 2e^{-} \rightarrow H_{2}$
Here,$n = 2$ and the reaction quotient $Q = \frac{p(H_{2})}{[H^{+}]^{2}}$.
Using the Nernst equation: $E_{H^{+}/H_{2}} = E^{0}_{H^{+}/H_{2}} - \frac{0.059}{n} \log Q$.
Since $E^{0}_{H^{+}/H_{2}} = 0 \ V$,we have $E_{H^{+}/H_{2}} = -\frac{0.059}{2} \log Q$.
For $E_{H^{+}/H_{2}}$ to be negative,$\log Q$ must be positive,which means $Q > 1$.
Evaluating the options:
$(A)$ $Q = \frac{1}{1^{2}} = 1$ (Not negative)
$(B)$ $Q = \frac{1}{2^{2}} = 0.25 < 1$ (Positive potential)
$(C)$ $Q = \frac{2}{1^{2}} = 2 > 1$ (Negative potential)
$(D)$ $Q = \frac{2}{2^{2}} = 0.5 < 1$ (Positive potential)
Therefore,the correct option is $(C)$.

(B) The given reactions represent the self-reduction (auto-reduction) process.
This process is used for the extraction of less reactive metals like $Cu$,$Hg$,and $Pb$ from their sulfide ores.
In the case of copper,the reactions are:
$2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$
$Cu_2S + 2Cu_2O \rightarrow 6Cu + SO_2$
Therefore,the metal $M$ is $Cu$.

(A) The hydrolysis of alkyl chlorides in aqueous $NaOH$ proceeds via $S_N1$ or $S_N2$ mechanisms.
$1$. Methoxymethyl chloride $(CH_3OCH_2Cl)$ undergoes $S_N1$ reaction very rapidly because the resulting carbocation $(CH_3OCH_2^+)$ is highly stabilized by resonance from the lone pair on the oxygen atom.
$2$. Benzyl chloride $(C_6H_5CH_2Cl)$ also forms a resonance-stabilized carbocation,but it is less stable than the methoxymethyl carbocation.
$3$. Neopentyl chloride $((CH_3)_3CCH_2Cl)$ is a primary alkyl halide with significant steric hindrance at the $\beta$-carbon,which makes $S_N2$ reactions extremely slow.
$4$. Propyl chloride $(CH_3CH_2CH_2Cl)$ is a primary alkyl halide that undergoes $S_N2$ reaction at a moderate rate.
Therefore,methoxymethyl chloride $(1)$ is hydrolysed most easily,and neopentyl chloride $(3)$ is hydrolysed most slowly.

(B) The boiling point depends on the strength of intermolecular forces. The order of strength of intermolecular forces is: $\text{Hydrogen bonding} > \text{Dipole-dipole interaction} > \text{Van der Waals forces}$.
$1$. $1$-butanol $(CH_3CH_2CH_2CH_2OH)$ exhibits intermolecular hydrogen bonding,which is the strongest force among the given compounds,resulting in the highest boiling point.
$2$. Butanone $(CH_3COCH_2CH_3)$ is a polar molecule and exhibits dipole-dipole interactions,which are stronger than Van der Waals forces but weaker than hydrogen bonding.
$3$. $n$-pentane and isopentane are non-polar alkanes and exhibit only Van der Waals forces. For alkanes,the boiling point decreases with branching because branching reduces the surface area of contact,leading to weaker Van der Waals forces. Thus,$n$-pentane has a higher boiling point than isopentane.
Therefore,the correct order of boiling points is: $\text{isopentane} < n\text{-pentane} < \text{butanone} < 1\text{-butanol}$.

(D) Conductivity depends on the number of ions and their ionic mobility.
$CH_{3}COOH$ is a weak electrolyte and dissociates poorly.
$NaCl$,$KNO_{3}$,and $HCl$ are strong electrolytes.
Among these,$HCl$ provides $H^{+}$ ions,which possess the highest ionic mobility in aqueous solution compared to $Na^{+}$,$K^{+}$,$Cl^{-}$,and $NO_{3}^{-}$ ions.
Therefore,$0.1 \ M \ HCl$ will have the highest conductivity.

(A) The reduction of nitrate ions $(NO_3^-)$ to ammonia $(NH_3)$ by zinc $(Zn)$ dust in an alkaline medium is a standard laboratory test for nitrates.
The chemical reaction is: $NaNO_3 + 4 Zn + 7 NaOH \rightarrow 4 Na_2ZnO_2 + NH_3 + 2 H_2O$.
Here,'$A$' represents the alkaline medium,which is $NaOH$ (caustic soda).

(B) The reaction of sodium sulfide with sulfur is given by: $Na_{2}S + nS \rightarrow Na_{2}S_{(n+1)}$,where $(X) = Na_{2}S_{(n+1)}$.
The reaction of sodium sulfite with sulfur is given by: $Na_{2}SO_{3} + S \rightarrow Na_{2}S_{2}O_{3}$,where $(Y) = Na_{2}S_{2}O_{3}$.

(D) For a body-centred cubic $(BCC)$ lattice, the number of atoms per unit cell is $Z = 2$.
The density formula is given by $\rho = \frac{Z \times M}{N_A \times a^3}$.
Given: $\rho = 5.0 \ g \ cm^{-3}$, $a = 200 \ pm = 200 \times 10^{-10} \ cm = 2 \times 10^{-8} \ cm$.
$5.0 = \frac{2 \times M}{6.022 \times 10^{23} \times (2 \times 10^{-8})^3}$.
$M = \frac{5.0 \times 6.022 \times 10^{23} \times 8 \times 10^{-24}}{2} = 12.044 \ g \ mol^{-1}$.
Number of moles in $100 \ g = \frac{100}{M} = \frac{100}{12.044} \approx 8.303 \ mol$.
Number of atoms $= \text{moles} \times N_A = \frac{100}{M} \times N_A$.
Substituting $M = \frac{5.0 \times N_A \times 8 \times 10^{-24}}{2}$:
Number of atoms $= \frac{100 \times 2}{5.0 \times 8 \times 10^{-24}} = \frac{200}{40 \times 10^{-24}} = 5 \times 10^{24}$.

(A) Given:
Mass of solute $(w_2)$ = $20 \ g$
Mass of solvent $(w_1)$ = $50 \ g$
Freezing point depression $(\Delta T_f)$ = $2 \ K$
$K_f$ of benzene = $1.72 \ K \ kg \ mol^{-1}$
Molar mass of naphthoic acid $(C_{11}H_8O_2)$ = $(11 \times 12) + (8 \times 1) + (2 \times 16) = 172 \ g \ mol^{-1}$
Using the formula:
$\Delta T_f = i \times K_f \times m$
$\Delta T_f = i \times K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$
$2 = i \times 1.72 \times \frac{20 \times 1000}{172 \times 50}$
$2 = i \times 1.72 \times \frac{20000}{8600}$
$2 = i \times 1.72 \times 2.3255...$
$2 = i \times 4$
$i = \frac{2}{4} = 0.5$

(A) Given:
Mole fraction of solute $(x_2)$ = $0.1$
Mole fraction of solvent $(x_1)$ = $1 - 0.1 = 0.9$
Density $(d)$ = $2.0 \ g \ cm^{-3}$
Molarity $(M)$ = Molality $(m)$
We know that:
$m = \frac{x_2 \times 1000}{x_1 \times M_1}$
$M = \frac{1000 \times d \times x_2}{x_1 M_1 + x_2 M_2}$
Since $M = m$:
$\frac{x_2 \times 1000}{x_1 \times M_1} = \frac{1000 \times d \times x_2}{x_1 M_1 + x_2 M_2}$
$\frac{1}{x_1 M_1} = \frac{d}{x_1 M_1 + x_2 M_2}$
$x_1 M_1 + x_2 M_2 = d \times x_1 M_1$
$0.9 M_1 + 0.1 M_2 = 2.0 \times 0.9 M_1$
$0.9 M_1 + 0.1 M_2 = 1.8 M_1$
$0.1 M_2 = 0.9 M_1$
$\frac{M_2}{M_1} = \frac{0.9}{0.1} = 9$
Therefore,the ratio of molecular weights of the solute and the solvent is $9$.

(B) The reaction is $^{14}N + {}_{2}^{4}He \rightarrow {}^{17}O + {}_{1}^{1}H$.
Since energy is absorbed,the mass of the products is greater than the mass of the reactants.
Mass defect $(\Delta m) = 0.00124 \ amu$.
$(m_{{}^{17}O} + m_{p}) - (m_{reactants}) = \Delta m$.
$m_{{}^{17}O} + 1.00782 \ amu - 18.00567 \ amu = 0.00124 \ amu$.
$m_{{}^{17}O} = 18.00567 + 0.00124 - 1.00782 = 16.99909 \ amu$.
Rounding to four decimal places,we get $16.9991 \ amu$.