MHT CET 2017 Chemistry Question Paper with Answer and Solution

(D) $(+)$ $2-$Methylbutan$-1-$ol and $(-)$ $2-$Methylbutan$-1-$ol are enantiomers of each other.
Enantiomers possess identical physical properties such as boiling point,density,and refractive index in an achiral environment.
However,they differ in their interaction with plane-polarized light,which is measured as specific rotation.
Therefore,they have different values for specific rotation.

(D) Food preservation involves techniques to prevent spoilage.
$Irradiation$,$addition \ of \ salts$,and $addition \ of \ heat$ (pasteurization/sterilization) are standard methods for food preservation.
$Hydration$ (adding water) actually promotes microbial growth and spoilage,whereas $dehydration$ (removing water) is used for preservation.
Therefore,$hydration$ is not a method used to preserve food.

(B) The electron gain enthalpy becomes more negative as we move across a period. However,in the case of halogens,Fluorine $(F)$ has a smaller size than Chlorine $(Cl)$.
Due to the small size of the $F$ atom,the incoming electron experiences significant inter-electronic repulsion from the electrons already present in the $2p$ subshell.
In contrast,the incoming electron in $Cl$ enters the $3p$ subshell,which is larger,resulting in less inter-electronic repulsion.
Therefore,$Cl$ has the highest value of negative electron gain enthalpy among the halogens.

(A) Alkaline $KMnO_4$ is known as Baeyer's reagent.
It is used as an oxidizing agent in organic chemistry to test for the presence of unsaturation (double or triple bonds) in organic compounds.

(A) In assigning $R-S$ configuration,the priority is determined by the Cahn-Ingold-Prelog $(CIP)$ sequence rules.
According to these rules,the atom with the higher atomic number attached directly to the chiral center is given higher priority.
Comparing the atoms attached to the chiral center: $S$ (atomic number $16$) in $-SO_3H$ has a higher atomic number than $C$ (atomic number $6$) in $-COOH$,$-CHO$,and $-C_6H_5$.
Therefore,the $-SO_3H$ group has the highest priority.
The decreasing order of priority is $-SO_3H > -COOH > -CHO > -C_6H_5$.

(C) In fullerene,each carbon atom is bonded to three other carbon atoms,forming a cage-like structure. Therefore,each carbon atom undergoes $sp^2$ hybridization.

(B) Pyrex glass is a type of borosilicate glass.
It is primarily composed of $60$ to $80 \%$ silica $(SiO_2)$,$10$ to $25 \%$ boron trioxide $(B_2O_3)$,and smaller amounts of aluminum oxide $(Al_2O_3)$ and other oxides.
Since $SiO_2$ constitutes the largest percentage,it is the chief constituent.

(A) The oxidizing power of an ion depends on its standard reduction potential $(E^\circ)$.
$Li^+$ has the most negative standard reduction potential $(E^\circ = -3.04 \ V)$ among the given alkali metal ions,which makes it the weakest oxidizing agent.
While $Li^+$ has a small ionic size and high hydration energy,its tendency to remain in the ionic state in aqueous solution is very high,making it extremely difficult to reduce to $Li(s)$.
Therefore,$Li^+$ is the weakest oxidizing agent among the choices provided.

(B) The balanced chemical equation is: $2 Na(s) + 2 C_2H_5OH(\ell) \rightarrow 2 C_2H_5O^-Na^+ + H_2(g) \uparrow$
From the stoichiometry,$2 \times 23 \ g$ of $Na$ produces $1 \text{ mole}$ of $H_2$ gas.
$46 \ g$ of $Na$ is equal to $2 \text{ moles}$ of $Na$.
Since $2 \times 23 \ g$ $(46 \ g)$ of $Na$ produces $1 \text{ mole}$ of $H_2$,the amount of $H_2$ produced is $1 \text{ mole}$.
Mass of $1 \text{ mole}$ of $H_2 = 2 \ g$.
Converting to $kg$: $2 \ g = 2 \times 10^{-3} \ kg$.

(A) The density of water is approximately $1 \ g/cm^3$. Thus,$20 \ cm^3$ of water has a mass of $20 \ g$.
The molar mass of water $(H_2O)$ is $18 \ g/mol$.
The number of moles of water is $n = \frac{20 \ g}{18 \ g/mol} = 1.11 \ mol$.
The number of molecules is $1.11 \times 6.022 \times 10^{23} \approx 6.69 \times 10^{23}$ molecules.
However,in liquid water,the molecules are closely packed. The actual volume occupied by the molecules themselves is negligible compared to the total volume of the liquid because the intermolecular space is very small. For standard calculations in this context,the volume of the molecules is considered to be the volume of the liquid itself,which is $20 \ cm^3$.

(C) The standard enthalpy of reaction is calculated using bond enthalpies as: $\Delta H^o_{\text{reaction}} = \sum H^o_{\text{reactant bonds}} - \sum H^o_{\text{product bonds}}$.
Option $C$ states the reverse of this relationship,making it the $INVALID$ equation.

(A) The first law of thermodynamics is given by the equation: $\Delta U = q + W$.
For an isothermal process,the temperature remains constant,which implies that the internal energy change is zero,i.e.,$\Delta U = 0$.
Substituting this into the first law equation,we get $0 = q + W$,which simplifies to $q = -W$.

(A) The formula for work done against constant external pressure is $W = -P_{\text{ext}} \Delta V$.
Given: $P_{\text{ext}} = 100 \ kPa = 100 \times 10^3 \ Pa$,$V_i = 1 \ m^3$,$V_f = 10 \ dm^3 = 0.01 \ m^3$.
Change in volume $\Delta V = V_f - V_i = 0.01 \ m^3 - 1 \ m^3 = -0.99 \ m^3$.
$W = -(100 \times 10^3 \ Pa) \times (-0.99 \ m^3) = 99,000 \ J = +99 \ kJ$.
Since the gas is being compressed,work is done on the system,hence the value is positive.

(C) The combustion reaction for ethane is: $C_2H_{6(g)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$.
Change in gaseous moles,$\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 3.5) = -2.5$.
The work done is given by $W = -\Delta n_g RT$.
For $1 \ mol$ of ethane: $W = -(-2.5 \times 8.314 \times 300) \ J = 6235.5 \ J = 6.2355 \ kJ$.
Molar mass of ethane $(C_2H_6)$ = $(2 \times 12) + (6 \times 1) = 30 \ g/mol = 30 \times 10^{-3} \ kg/mol$.
Given mass of ethane = $9 \times 10^{-2} \ kg = 90 \ g$.
Number of moles of ethane = $\frac{90 \ g}{30 \ g/mol} = 3 \ mol$.
Total work done = $3 \ mol \times 6.2355 \ kJ/mol = 18.7065 \ kJ \approx 18.71 \ kJ$.

 $l_x=40 \ cm, l_R=60 \ cm$
$\frac{x}{R}=\frac{l_x}{l_R}=\frac{40}{60}=\frac{2}{3} \ldots \ldots . .(i)$
$\therefore \frac{x+30}{R}=\frac{60}{40}=\frac{3}{2}$
$\therefore \frac{x+30}{R}=\frac{3}{2}$
$\therefore 2 \times(+30)=3 R$
$R=\frac{2(x+30)}{3} \ldots \ldots . .(ii)$
From $(i)$ and $(ii),$
$\frac{3 x}{2(x+30)}=\frac{2}{3}$
$9 x=4 x+120$
$5 x=120 \Rightarrow x=24 \Omega$

(C) The magnetic induction $B$ inside a long solenoid is given by $B = \frac{\mu_0 N I}{L}$.
Magnetic flux $\phi$ through the cross-sectional area $A$ is $\phi = B A = \frac{\mu_0 N I A}{L}$.
The magnetic moment $M$ of the solenoid is defined as $M = N I A$.
Substituting $M$ into the flux equation,we get $\phi = \frac{\mu_0 M}{L}$.
Rearranging for $M$,we have $M = \frac{\phi L}{\mu_0}$.
Given $\phi = 1.57 \times 10^{-6} \ Wb$,$L = 0.6 \ m$,and $\mu_0 = 4\pi \times 10^{-7} \ T \ m/A$.
$M = \frac{1.57 \times 10^{-6} \times 0.6}{4 \times 3.14 \times 10^{-7}}$.
$M = \frac{1.57 \times 10^{-6} \times 0.6}{12.56 \times 10^{-7}} = \frac{0.942 \times 10^{-6}}{1.256 \times 10^{-6}} = 0.75 \ A \ m^2$.

(D) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains constant,the volume of the big drop equals the sum of the volumes of $n$ small drops:
$\frac{4}{3} \pi R^3 = n \cdot \frac{4}{3} \pi r^3$
$R^3 = n r^3 \implies R = n^{1/3} r$
Surface energy $E$ is given by $E = T \cdot A$,where $T$ is surface tension and $A$ is the surface area.
Surface energy of $n$ small drops $(E_2)$ = $n \times (4 \pi r^2 T) = 4 n \pi r^2 T$
Surface energy of the big drop $(E_1)$ = $4 \pi R^2 T$
Ratio $\frac{E_2}{E_1} = \frac{4 n \pi r^2 T}{4 \pi R^2 T} = \frac{n r^2}{R^2}$
Substituting $R = n^{1/3} r$:
$\frac{E_2}{E_1} = \frac{n r^2}{(n^{1/3} r)^2} = \frac{n r^2}{n^{2/3} r^2} = n^{1 - 2/3} = n^{1/3}$
Thus,the ratio is $\sqrt[3]{n}: 1$.

(A) The upward force due to surface tension is given by $F = T \times L$,where $L$ is the inner circumference of the capillary tube $(L = 2 \pi r)$.
Given: $F = 105 \text{ dyne} = 105 \times 10^{-5} \text{ N}$ and $T = 7 \times 10^{-2} \text{ N m}^{-1}$.
Equating the forces: $F = T \times L$.
$L = \frac{F}{T} = \frac{105 \times 10^{-5} \text{ N}}{7 \times 10^{-2} \text{ N m}^{-1}}$.
$L = 15 \times 10^{-3} \text{ m} = 1.5 \times 10^{-2} \text{ m}$.
Converting to centimeters: $L = 1.5 \text{ cm}$.

(A) Let the displacement from the mean position be $x_1$ when velocity is $u$ and acceleration is $\alpha$. Since acceleration is $\alpha = \omega^2 x_1$,we have $x_1 = \frac{\alpha}{\omega^2}$.
At this position,$u^2 = \omega^2(A^2 - x_1^2) = \omega^2 A^2 - \omega^2 x_1^2 = \omega^2 A^2 - \alpha x_1$.
Let the displacement be $x_2$ when velocity is $v$ and acceleration is $\beta$. Since acceleration is $\beta = \omega^2 x_2$,we have $x_2 = \frac{\beta}{\omega^2}$.
At this position,$v^2 = \omega^2(A^2 - x_2^2) = \omega^2 A^2 - \omega^2 x_2^2 = \omega^2 A^2 - \beta x_2$.
Subtracting the two velocity equations: $u^2 - v^2 = (\omega^2 A^2 - \alpha x_1) - (\omega^2 A^2 - \beta x_2) = \beta x_2 - \alpha x_1$.
Substituting $x_1 = \frac{\alpha}{\omega^2}$ and $x_2 = \frac{\beta}{\omega^2}$:
$u^2 - v^2 = \beta(\frac{\beta}{\omega^2}) - \alpha(\frac{\alpha}{\omega^2}) = \frac{\beta^2 - \alpha^2}{\omega^2}$.
Thus,$\omega^2 = \frac{\beta^2 - \alpha^2}{u^2 - v^2}$.
The distance between the two positions is $|x_2 - x_1| = |\frac{\beta}{\omega^2} - \frac{\alpha}{\omega^2}| = \frac{\beta - \alpha}{\omega^2}$.
Substituting $\omega^2$: $|x_2 - x_1| = (\beta - \alpha) \cdot \frac{u^2 - v^2}{\beta^2 - \alpha^2} = \frac{(\beta - \alpha)(u^2 - v^2)}{(\beta - \alpha)(\beta + \alpha)} = \frac{u^2 - v^2}{\alpha + \beta}$.

(D) The displacement of a particle performing $SHM$ starting from the extreme position is given by $x(t) = A \cos(\omega t)$.
The acceleration of the particle is given by $a(t) = \frac{d^2x}{dt^2} = -A\omega^2 \cos(\omega t)$.
We can rewrite the acceleration as $a(t) = A\omega^2 \cos(\omega t + \pi)$.
Comparing the phase of displacement $(\omega t)$ and acceleration $(\omega t + \pi)$, the phase difference is $\pi$ rad.

(C) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $P = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature.
For a spherical body,the surface area $A = 4 \pi r^2$.
Thus,$P = \sigma (4 \pi r^2) T^4$.
Given that both bodies radiate the same power,we have $P_1 = P_2$.
Therefore,$\sigma (4 \pi r_1^2) T_1^4 = \sigma (4 \pi r_2^2) T_2^4$.
Simplifying the equation,we get $r_1^2 T_1^4 = r_2^2 T_2^4$.
Rearranging to find the ratio $\frac{r_2}{r_1}$,we get $\frac{r_2^2}{r_1^2} = \frac{T_1^4}{T_2^4}$.
Taking the square root on both sides,we get $\frac{r_2}{r_1} = \sqrt{\frac{T_1^4}{T_2^4}} = \left(\frac{T_1}{T_2}\right)^2$.

(C) For isomeric alcohols,the boiling point decreases as the extent of branching increases.
$n$-butyl alcohol $(CH_3CH_2CH_2CH_2OH)$ is a straight-chain alcohol with the highest surface area and strongest intermolecular hydrogen bonding.
tert-butyl alcohol $((CH_3)_3COH)$ is the most branched isomer,which reduces the surface area available for van der Waals interactions,resulting in the lowest boiling point.

(D) The reaction of $Tert$-butyl methyl ether with $HI$ in cold conditions follows an $S_N1$ mechanism.
Since the $Tert$-butyl group can form a stable $Tert$-butyl carbocation,the $C-O$ bond between the $Tert$-butyl group and the oxygen atom breaks.
The reaction proceeds as follows:
$(CH_3)_3C-O-CH_3 + HI \xrightarrow{\text{cold}} (CH_3)_3C-I + CH_3OH$
Thus,the products formed are $Tert$-butyl iodide and methanol (methyl alcohol).

(B) The reaction of aldehydes or ketones with phenylhydrazine $(C_6H_5-NH-NH_2)$ involves the nucleophilic addition of the amine group to the carbonyl carbon followed by the elimination of a water molecule.
This condensation reaction results in the formation of a phenylhydrazone.
The general reaction is:
$R_2C=O + H_2N-NH-C_6H_5 \rightarrow R_2C=N-NH-C_6H_5 + H_2O$
Thus,the product formed is a phenylhydrazone.

(C) The nitration of $Urotropine$ (hexamethylenetetramine) with concentrated nitric acid $(HNO_3)$ and ammonium nitrate yields $Cyclonite$,also known as $RDX$ (Research Department Explosive).

(A) The iodoform test is given by compounds containing the $CH_3CO-$ group or the $CH_3CH(OH)-$ group.
Isopropyl alcohol $(CH_3CH(OH)CH_3)$ contains the $CH_3CH(OH)-$ group,so it gives a positive iodoform test.
Propionaldehyde $(CH_3CH_2CHO)$ does not contain the $CH_3CO-$ group.
Ethylphenyl ketone $(C_6H_5COCH_2CH_3)$ does not contain the $CH_3CO-$ group.
Benzyl alcohol $(C_6H_5CH_2OH)$ does not contain the $CH_3CH(OH)-$ group.

(A) The reaction with $p$-toluene sulphonyl chloride (Hinsberg reagent) is used to distinguish between $1^\circ$,$2^\circ$,and $3^\circ$ amines.
$1^\circ$ amines $(R-NH_2)$ react to form an $N$-alkyl $p$-toluene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen. This makes the product soluble in alkali (forming a clear solution).
Upon acidification of this clear solution,the $N$-alkyl $p$-toluene sulphonamide precipitates out as an insoluble compound.
$2^\circ$ amines form a product that is insoluble in alkali,and $3^\circ$ amines do not react with the reagent.
Therefore,the primary amine $C_2H_5NH_2$ is the correct answer.

(B) The Hofmann bromamide degradation reaction involves the treatment of an amide with bromine in an aqueous or alcoholic solution of a strong base like $KOH$ or $NaOH$.
This reaction results in the formation of a primary amine $(R-NH_2)$ with one carbon atom less than the original amide $(R-CONH_2)$.
Therefore,the statement that it gives a tertiary amine is incorrect.

(D) The basic strength of aniline is determined by the availability of the lone pair of electrons on the nitrogen atom.
Electron-withdrawing groups $(EWG)$ decrease the electron density on the nitrogen atom,thereby decreasing the basic strength.
Among the given options,$-OCH_3$,$-CH_3$,and $-NH_2$ are electron-donating groups $(EDG)$ due to their $+M$ or $+I$ effects,which increase the basic strength.
$-C_6H_5$ (phenyl group) acts as an electron-withdrawing group due to its $-I$ effect and resonance,thus decreasing the basic strength of aniline.

(D) The chemical formula of Stachyose is $C_{24}H_{42}O_{21}$.
Therefore,a single molecule of Stachyose contains $24$ carbon atoms.

(A) An amino acid is basic if it contains more amino groups than carboxyl groups in its side chain.
Histidine contains an imidazole ring which acts as a base.
Therefore,Histidine is basic in nature.

(C) $D-2-\text{deoxyribose}$ sugar molecule is present in $DNA$.

(D) The structures of the given acids are as follows:
$1$. Phthalic acid: $C_6H_4(COOH)_2$ (contains two $-COOH$ groups).
$2$. Adipic acid: $HOOC(CH_2)_4COOH$ (contains two $-COOH$ groups).
$3$. Glutaric acid: $HOOC(CH_2)_3COOH$ (contains two $-COOH$ groups).
$4$. Salicylic acid: $C_6H_4(OH)(COOH)$ (contains both $-OH$ and $-COOH$ groups).
Therefore,Salicylic acid is the correct answer.

(B) Due to the small size of $N$,it has a unique ability to form $P \pi-P \pi$ multiple bonds.
Other elements in the group have larger atomic sizes and diffuse orbitals,which makes effective sidewise overlapping difficult,thus they prefer to form single bonds.

(A) The reaction for the inversion of cane sugar is: $C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$.
In this reaction,$H_2O$ is present in large excess,so its concentration remains effectively constant during the reaction.
Therefore,the rate of the reaction depends only on the concentration of sucrose,making it a pseudo-first order reaction.

(A) For a first order reaction,the half-life $(t_{1/2})$ is independent of the initial concentration of the reactant.
The formula for half-life is given by: $t_{1/2} = \frac{0.693}{k}$
Given,$k = 7.0 \times 10^{-4} \ s^{-1}$
Substituting the value of $k$ in the formula:
$t_{1/2} = \frac{0.693}{7.0 \times 10^{-4} \ s^{-1}}$
$t_{1/2} = 990 \ s$

(C) The Arrhenius equation is given by the expression: $K = A \cdot e^{-E_{a} / (RT)}$.
This can be rewritten as: $K = \frac{A}{e^{E_{a} / (RT)}}$.
Therefore,the correct representation among the given options is $K = \frac{A}{e^{E_{a} / (RT)}}$.

(B) Bithional is added to soaps to impart antiseptic properties and is used to reduce the odours produced by bacterial decomposition of organic matter on the skin.

(D) The $EAN$ (Effective Atomic Number) rule states that the total number of electrons around the central metal ion in a complex should equal the atomic number of the next noble gas.
For $[Pt(NH_3)_6]^{4+}$: $EAN = 78 - 4 + 6(2) = 86$ (matches Radon).
For $[Fe(CN)_6]^{4-}$: $EAN = 26 - 2 + 6(2) = 36$ (matches Krypton).
For $[Zn(NH_3)_4]^{2+}$: $EAN = 30 - 2 + 4(2) = 36$ (matches Krypton).
For $[Cu(NH_3)_4]^{2+}$: $EAN = 29 - 2 + 4(2) = 35$ (does not match Krypton,$36$).
Thus,$[Cu(NH_3)_4]^{2+}$ is an exception to the $EAN$ rule.

(B) The complex $K_3[Fe(CN)_6]$ dissociates as $3K^+ + [Fe(CN)_6]^{3-}$.
Let the oxidation state of $Fe$ be $x$.
In the complex ion $[Fe(CN)_6]^{3-}$,the charge on $CN^-$ is $-1$.
So,$x + 6(-1) = -3$.
$x - 6 = -3 \Rightarrow x = +3$.
The coordination number is the number of ligand donor atoms bonded to the central metal ion. Since $6$ $CN^-$ ligands are attached,the coordination number is $6$.

(A) The oxidation state of $Mn$ in $MnO_4^{2-}$ is $+6$.
The electronic configuration of $Mn^{6+}$ is $[Ar] 3d^1$.
Due to the presence of one unpaired electron in the $3d$ orbital,the $MnO_4^{2-}$ ion is paramagnetic.
The manganate ion $(MnO_4^{2-})$ exhibits a green colour.

(C) The lead chamber process involves the oxidation of $SO_2$ to $SO_3$ using air in the presence of nitrogen oxides as a catalyst.
Specifically,$NO$ (nitric oxide) acts as the catalyst in this process.

(D) $Zr$ $(Z=40)$ belongs to the $5^{th}$ period,while $Hf$ $(Z=72)$ belongs to the $6^{th}$ period.
Due to lanthanoid contraction,$Zr$ and $Hf$ exhibit nearly identical atomic and ionic radii,making them chemical twins.
Since they belong to different periods ($5^{th}$ and $6^{th}$),the statement that they belong to the same period is incorrect.

(D) In a lead storage battery (lead accumulator),the positive electrode is made of $PbO_2$.
During discharge (when used as a source of electrical energy),the reduction reaction at the positive electrode is:
$PbO_{2\text{(s)}} + 4H^{+}{_{\text{(aq)}}} + SO_4^{2-}{_{\text{(aq)}}} + 2e^{-} \rightarrow PbSO_{4\text{(s)}} + 2H_2O_{\text{(l)}}$
In this reaction,the oxidation state of lead changes from $+4$ in $PbO_2$ to $+2$ in $PbSO_4$.
Therefore,the reduction process is represented by $Pb^{4+} \rightarrow Pb^{2+}$.

(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\rho)$.
Since the $SI$ unit of resistivity is $\Omega \ m$,the unit of conductivity is $\Omega^{-1} \ m^{-1}$.
Since $\Omega^{-1}$ is represented as Siemens $(S)$,the $SI$ unit of conductivity is $S \ m^{-1}$.

(B) Magnesite is $MgCO_3$.
It is a mineral of Magnesium,not iron.
Haematite $(Fe_2O_3)$,Magnetite $(Fe_3O_4)$,and Siderite $(FeCO_3)$ are minerals of iron.

(C) Hoope's process is the electrolytic refining process employed to purify aluminium.

(B) The reaction of alkyl halides (like ethyl bromide) with sodium iodide $(NaI)$ in the presence of dry acetone to form alkyl iodides (like ethyl iodide) is a classic example of the Finkelstein reaction.
The reaction is: $C_2H_5Br + NaI \xrightarrow{\text{dry acetone}} C_2H_5I + NaBr$.

(B) To find the oxidation state of phosphorus $(x)$ in $H_3PO_2$ (Phosphinic acid):
$3(+1) + x + 2(-2) = 0$
$3 + x - 4 = 0$
$x - 1 = 0$
$x = +1$
Therefore,phosphorus is in the $+1$ oxidation state in Phosphinic acid $(H_3PO_2)$.

(A) Argon $(Ar)$ is a noble gas belonging to Group $18$.
Noble gases exist as monoatomic gases because they have a stable valence shell configuration $(ns^2 np^6)$.
Oxygen $(O_2)$,Nitrogen $(N_2)$,and Bromine $(Br_2)$ exist as diatomic molecules in their standard states.

(A) Bakelite is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde. Because it is a poor conductor of heat and electricity and is resistant to heat,it is widely used for making handles of cookers,frying pans,and electrical switches.

(A) Dextron (also known as Dexon) is a biodegradable polymer formed by the copolymerization of lactic acid $(CH_3CH(OH)COOH)$ and glycolic acid $(HOCH_2COOH)$.
This reaction involves the elimination of water molecules to form an ester linkage,resulting in a polyester copolymer.
The chemical reaction is:
$n \ CH_3-CH(OH)-COOH + n \ HO-CH_2-COOH \rightarrow -[O-CH(CH_3)-CO-O-CH_2-CO]_n- + 2n \ H_2O$
Thus,the correct monomers are lactic acid and glycolic acid.

(A) Due to the high $H-O$ bond dissociation enthalpy,$H_2O$ does not act as a reducing agent.
In contrast,$H_2S$,$H_2Se$,and $H_2Te$ act as reducing agents.
The reducing character increases in the order: $H_2S < H_2Se < H_2Te$.
This increase in reducing power is due to the decrease in $H-E$ bond strength as the size of the central atom $E$ increases down the group.

(D) The solubility of most alkali metal salts increases significantly with temperature,but $NaBr$ shows a relatively small change in solubility compared to others like $KNO_3$ or $NaNO_3$ within certain temperature ranges. This is due to the specific lattice energy and hydration energy balance of $NaBr$.

(D) The Van't Hoff factor $(i)$ represents the number of particles a solute dissociates into in a solution.
$Urea$ is a non-electrolyte,so it does not dissociate,and its $(i)$ value is $1$.
$AlCl_3$ dissociates as $AlCl_3 \rightarrow Al^{3+} + 3Cl^-$,so $(i) = 4$.
$K_2SO_4$ dissociates as $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$,so $(i) = 3$.
$NH_4Cl$ dissociates as $NH_4Cl \rightarrow NH_4^+ + Cl^-$,so $(i) = 2$.
Therefore,$Urea$ has the lowest Van't Hoff factor.

(A) The formula for osmotic pressure is $\pi = CRT$.
First,calculate the number of moles of cane sugar: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{34.2 \ g}{342 \ g \ mol^{-1}} = 0.1 \ mol$.
The concentration $C = \frac{n}{V} = \frac{0.1 \ mol}{1 \ L} = 0.1 \ mol \ L^{-1}$.
The temperature $T = 20 + 273 = 293 \ K$.
Substituting the values: $\pi = 0.1 \ mol \ L^{-1} \times 0.082 \ L \ atm \ K^{-1} \ mol^{-1} \times 293 \ K = 2.4026 \ atm \approx 2.40 \ atm$.

(A) The formula for molality $(m)$ is given by: $m = \frac{w_2}{M_2} \times \frac{1000}{w_1(g)}$
Where $w_2 = 15.20 \ g$ (mass of solute),$M_2 = 60 \ g \ mol^{-1}$ (molar mass of urea),and $w_1 = 150 \ g$ (mass of solvent).
Substituting the values:
$m = \frac{15.20}{60} \times \frac{1000}{150}$
$m = 0.2533 \times 6.6667$
$m = 1.689 \ mol \ kg^{-1}$