GUJCET 2020 Chemistry Question Paper with Answer and Solution

(B) The reaction of a metal '$M$' with an alcohol $(ROH)$ can be represented as:
$M + nROH \rightarrow M(OR)_n + \frac{n}{2} H_2$
Given that $1 \ mol$ of '$M$' produces $1.5 \ mol$ of $H_2$,we can equate the stoichiometric coefficient of $H_2$ to $1.5$:
$\frac{n}{2} = 1.5$
$n = 1.5 \times 2 = 3$
Therefore,the valency of the metal '$M$' is $3$.

(D) The power $P$ of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Given:
Power $P = 100 \ W$
Voltage $V = 220 \ V$
Rearranging the formula to solve for resistance $R$:
$R = \frac{V^2}{P}$
Substituting the given values:
$R = \frac{(220)^2}{100}$
$R = \frac{48400}{100}$
$R = 484 \ \Omega$
Therefore,the resistance of the bulb is $484 \ \Omega$.

(D) The given reaction is a Wurtz reaction,where two molecules of an alkyl halide $(R'-Cl)$ react with sodium in the presence of dry ether to form a symmetric alkane.
The reaction is: $2R'-Cl + 2Na \xrightarrow{\text{Dry ether}} R'-R' + 2NaCl$.
The product formed is $2,3-\text{dimethylbutane}$,which has the structure $CH_3-CH(CH_3)-CH(CH_3)-CH_3$.
By splitting this symmetric alkane at the center,we get two identical fragments of $CH_3-CH(CH_3)-$. Thus,the alkyl group $R'$ is the isopropyl group,$CH_3-CH(CH_3)-$.
Therefore,$R'-Cl$ is isopropyl chloride.

(B) The molecular formula $C_4H_9Br$ corresponds to butyl bromide isomers.
Among the structural isomers,$2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ contains a chiral carbon atom (the $C_2$ carbon is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-Br$).
$A$ molecule with one chiral center exists as a pair of enantiomers (two optically active isomers: $d$- and $l$-forms).
Therefore,there are $2$ optically active isomers possible for this compound.

(A) The boiling point of organic compounds depends on the intermolecular forces present.
$1$. Pentan-$1$-ol is an alcohol,which exhibits strong intermolecular hydrogen bonding.
$2$. Pentanal is an aldehyde,which has dipole-dipole interactions.
$3$. Ethoxy ethane is an ether,which has weaker dipole-dipole interactions.
$4$. $n$-Butane is an alkane,which only has weak London dispersion forces.
Since hydrogen bonding is the strongest intermolecular force among these,Pentan-$1$-ol has the highest boiling point.

(D) Step $1$: Reaction of Grignard reagent with $CO_2$.
$C_6H_5CH_2MgBr + CO_2$ $\rightarrow C_6H_5CH_2COOMgBr$ $\xrightarrow{H_3O^+} C_6H_5CH_2COOH$ (Phenylacetic acid,'$X$').
Step $2$: Decarboxylation of the carboxylic acid.
$C_6H_5CH_2COOH \xrightarrow{NaOH + CaO, \Delta} C_6H_5CH_3$ (Toluene,'$Y$').
The final product '$Y$' is $C_6H_5CH_3$.

(A) The $pK_{a}$ value is inversely proportional to the acid strength. Stronger acids have lower $pK_{a}$ values,while weaker acids have higher $pK_{a}$ values.
Electron-withdrawing groups (EWGs) increase the acidity of carboxylic acids by stabilizing the carboxylate anion through the inductive effect.
Comparing the substituents attached to the $CH_{2}COOH$ group:
$1$. $C_{6}H_{5}-$ (Phenyl group): Weakly electron-withdrawing or electron-donating by resonance,making it the least acidic among the options.
$2$. $O_{2}N-$ (Nitro group): Strong electron-withdrawing group ($-I$ and $-M$ effect).
$3$. $F-$ (Fluoro group): Strong electron-withdrawing group ($-I$ effect).
$4$. $NC-$ (Cyano group): Strong electron-withdrawing group ($-I$ effect).
Since $C_{6}H_{5}CH_{2}COOH$ has the weakest electron-withdrawing group,it is the least acidic and therefore has the highest $pK_{a}$ value.

(C) The oxidation of secondary alcohols like cyclohexanol to ketones like cyclohexanone can be achieved using various oxidizing agents.
Anhydrous $CrO_3$ (Jones reagent or similar chromium-based oxidants) is a standard reagent used for the oxidation of secondary alcohols to ketones.
Therefore,the correct reagent is anhydrous $CrO_3$.

(C) Scurvy is a disease caused by the deficiency of Vitamin $C$.
Vitamin $C$ is chemically known as $Ascorbic \ acid$.
Therefore,the correct option is $C$.

(B) $(i)$ Most naturally occurring amino acids have $L$-configuration. This is $True$ $(T)$.
$(ii)$ $\beta-D$-ribose sugar is the sugar component present in $RNA$. This is $True$ $(T)$.
$(iii)$ Amylose is a water-insoluble component of starch,composed of long unbranched chains of $\alpha-D^+$-glucose units linked by $\alpha$-glycosidic linkages. This is $True$ $(T)$.
$(iv)$ All monosaccharides,whether aldoses or ketoses,are reducing sugars because they contain a free aldehyde or ketone group. Therefore,the statement that all monosaccharides are non-reducing is $False$ $(F)$.
Thus,the sequence is $T, T, T, F$. However,based on the provided options,the closest match for the logic is $TTFF$ (Option $B$),assuming a potential typo in the source question regarding the solubility or classification of amylose in the provided options.

(B) For a first-order reaction,the integrated rate equation is $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,$m = \frac{k}{2.303}$,and $c = 0$.
Since the y-intercept $c$ is $0$,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ passes through the origin.
Therefore,the correct option is $B$.

(A) For a first order reaction,the half-life is given by $t_{1/2} = \frac{0.693}{k}$.
Given $t_{1/2} = 40 \ minutes$.
Convert time to seconds: $t_{1/2} = 40 \times 60 \ s = 2400 \ s$.
Now,calculate the rate constant $k = \frac{0.693}{t_{1/2}}$.
$k = \frac{0.693}{2400} \ s^{-1} = 0.00028875 \ s^{-1}$.
Rounding to three significant figures,$k = 2.88 \times 10^{-4} \ s^{-1}$.

(B) For a first-order reaction,the integrated rate equation is $\log \frac{[R]_0}{[R]} = \frac{kt}{2.303}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \log \frac{[R]_0}{[R]}$,$x = t$,$m = \frac{k}{2.303}$,and $c = 0$.
Since the intercept $c$ is $0$,the graph of $\log \frac{[R]_0}{[R]}$ versus $t$ passes through the origin.
Therefore,the correct option is $B$.

(D) Electrical conductivity in an aqueous solution depends on the number of ions produced upon dissociation.
$1$. $[Co(H_2O)_4Cl_2]Cl \rightarrow [Co(H_2O)_4Cl_2]^+ + Cl^-$ (Produces $2$ ions)
$2$. $[Co(H_2O)_3Cl_3]$ (Does not dissociate,$0$ ions)
$3$. $[Co(H_2O)_5Cl]Cl_2 \rightarrow [Co(H_2O)_5Cl]^{2+} + 2Cl^-$ (Produces $3$ ions)
$4$. $[Co(H_2O)_6]Cl_3 \rightarrow [Co(H_2O)_6]^{3+} + 3Cl^-$ (Produces $4$ ions)
Since $[Co(H_2O)_6]Cl_3$ produces the maximum number of ions ($4$ ions),it exhibits the highest electrical conductivity.

(C) The wavelength of light absorbed $(\lambda)$ is inversely proportional to the crystal field splitting energy $(\Delta_o)$.
$\lambda \propto \frac{1}{\Delta_o}$.
To absorb the maximum wavelength,the complex must have the minimum crystal field splitting energy $(\Delta_o)$.
According to the spectrochemical series,the strength of ligands is: $CN^{-} > NH_3 > H_2O > Cl^{-}$.
Among the given ligands,$Cl^{-}$ is the weakest ligand,which results in the smallest $\Delta_o$ value.
Therefore,the complex $[CoCl(NH_3)_5]^{2+}$ will have the minimum $\Delta_o$ and will absorb the maximum wavelength of light.

(B) The complex is $[Co(NH_3)_6][Cr(CN)_6]$.
This complex consists of both a cationic part $[Co(NH_3)_6]^{3+}$ and an anionic part $[Cr(CN)_6]^{3-}$.
Coordination isomerism occurs in complexes where both the cation and anion are complex ions,and the ligands can be exchanged between the metal centers.
Therefore,the isomer $[Cr(NH_3)_6][Co(CN)_6]$ is possible.
Thus,the correct answer is $B$.

(A) The similarity in the properties of $Zr$ and $Hf$ is primarily due to the lanthanoid contraction. Due to the filling of $4f$ orbitals in the elements between $La$ and $Hf$, the atomic radius of $Hf$ $(159 \text{ pm})$ is almost identical to that of $Zr$ $(160 \text{ pm})$. This phenomenon is known as the lanthanoid contraction, which results in similar atomic and ionic radii for elements of the same group in the $4d$ and $5d$ series. Therefore, the correct answer is $A$.

(D) The magnetic moment $(\mu)$ is given by the formula $\mu = \sqrt{n(n+2)} \ BM$,where $n$ is the number of unpaired electrons.
Given $\mu = 5.92 \ BM$,we have $\sqrt{n(n+2)} = 5.92$,which implies $n \approx 5$.
For a divalent ion $(M^{2+})$ to have $5$ unpaired electrons,the electronic configuration must be $3d^5$.
Among the given elements,$Mn$ $(Z=25)$ has the ground state configuration $[Ar] 3d^5 4s^2$.
Upon forming the $Mn^{2+}$ ion,it loses two $4s$ electrons,resulting in the configuration $[Ar] 3d^5$.
This configuration contains $5$ unpaired electrons,consistent with the magnetic moment of $5.92 \ BM$.

(C) According to Faraday's law of electrolysis,the mass deposited $(w)$ is given by $w = \frac{I \times t \times M_{atomic}}{n \times F}$,where $I = 1.5 \ A$,$t = 10 \times 60 = 600 \ s$,$M_{atomic} = 63 \ g/mol$,and $F = 96500 \ C/mol$.
Substituting the values: $0.2938 = \frac{1.5 \times 600 \times 63}{n \times 96500}$.
$n = \frac{1.5 \times 600 \times 63}{0.2938 \times 96500} \approx \frac{56700}{28351.7} \approx 2$.
Since the valency of the metal $M$ is $2$,the metal ion is $M^{2+}$.
Therefore,the formula of the metal halide is $MCl_2$.

(B) The variation of molar conductivity $(\Lambda_m)$ with concentration $(C)$ for strong electrolytes is given by the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$.
In this equation,$A$ is a constant,and the slope of the graph of $\Lambda_m$ versus $\sqrt{C}$ is $-A$,which is negative.
Among the given options,$CH_3COONa$ (Sodium acetate) is a strong electrolyte,whereas $NH_4OH$,$CH_3COOH$,and $H_2O$ are weak electrolytes.
Therefore,the graph for Sodium acetate shows a negative slope.

(D) The strength of a reducing agent is inversely proportional to its standard reduction potential $(E^0_{red})$.
$A$ lower (more negative) reduction potential indicates that the substance is more easily oxidized and is therefore a stronger reducing agent.
Comparing the given values:
$1. E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$
$2. E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$
$3. E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$
$4. E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
Since $E^0_{Zn^{2+}/Zn}$ has the lowest value $(-0.76 \text{ V})$,$Zn$ is the strongest reducing agent.

(A) The osmotic pressure $\pi$ is given by the formula $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature.
For $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$,the degree of dissociation $\alpha$ is related to $[H^+]$.
Given $pH = 2$,so $[H^+] = 10^{-pH} = 10^{-2} = 0.01 \ M$.
Since $[H^+] = C \alpha$,we have $0.01 = 0.5 \times \alpha$,which gives $\alpha = 0.01 / 0.5 = 0.02$.
The van't Hoff factor $i = 1 + \alpha = 1 + 0.02 = 1.02$.
Substituting the values into the osmotic pressure formula: $\pi = 1.02 \times 0.5 \times RT = 0.51 \ RT$.

(C) Given,mole fraction of solute $(x_B)$ = $0.25$.
Since it is an aqueous solution,the mole fraction of solvent (water,$x_A$) = $1 - 0.25 = 0.75$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{n_B}{W_A \text{ (in kg)}} = \frac{n_B}{n_A \times M_A \times 10^{-3}}$,where $M_A$ is the molar mass of water $(18 \ g/mol)$.
$m = \frac{x_B}{x_A \times M_A \text{ (in kg/mol)}} = \frac{0.25}{0.75 \times 18 \times 10^{-3}}$.
$m = \frac{0.25}{0.0135} \approx 18.52 \ m$.

(D) The solubility of a solid solute in a liquid solvent depends on the nature of the solute and solvent (like dissolves like) and the temperature.
However,solids are incompressible,so the effect of pressure on the solubility of a solid in a liquid is negligible.
Therefore,the maximum amount of solid solute that can be dissolved does not depend upon pressure $(iii)$.
Thus,the correct option is $D$.