VITEEE 2024 Question Paper with Solutions

$\therefore$ RSUYC is different.

The pattern of the series is as shown below.

$\therefore ?=935$

As,

Similarly,

Pattern of the series is as follow.

$\therefore ?=\mathrm{VNU}$

As,

$=67-2$ (Number of vowel)

$=65$

and,

$=72-2$ (Number of vowel)

$=70$

Similarly

$=93-1$ (Number of vowel)

$=92$

$\frac{\text{ Total sale of Branch }{𝐵}_2}{\text{ Total sale of Branch }{𝐵}_4}=\frac{75+65}{85+95}$

Required ratio $=\frac{140}{180}=\frac{7}{9}$

According to the question, Average sales of all the five branches (in thousand number) for the year 2010

$\begin{array}{cc}& =\frac{80+75+95+85+75}{5}\\ & =\frac{410}{5}=82\end{array}$

According to the question,

Conclusion I and II are complementary pair. Hence, option (d) is correct.

Let's analyze the information given in each statement to see if we have enough to determine Ritika’s birth year.

Statement I: “Ritika at present is 25 years younger to her mother.”

This tells us that the age difference between Ritika and her mother is 25 years.

In equation form, if we let the mother’s birth year be $𝑀$ and Ritika’s birth year be $𝑅$, then:

$𝑅=𝑀+25$

However, without knowing the actual birth year of the mother (or any specific year), we cannot determine $𝑅$ uniquely.

Thus, Statement I by itself is not sufficient.

Statement II: “Ritika's sister who was born in 1964, is 35 years younger to her mother.”

This statement allows us to determine the mother’s birth year.

If the sister was born in 1964 and is 35 years younger than her mother, then their birth years differ by 35 years:

$𝑀=1964-35=1929$

Now we know the mother was born in 1929.

However, Statement II does not tell us anything about Ritika’s age or her relation to the mother.

Without knowing the 25-year difference between Ritika and her mother (as given in Statement I), we cannot find Ritika’s birth year based solely on Statement II.

Therefore, Statement II by itself is not sufficient.

Combine Statements I and II:

From Statement II, we already calculated the mother’s birth year as 1929.

Now, using Statement I, where Ritika is 25 years younger than her mother, we find:

$𝑅=1929+25=1954$

With both pieces of information, we can uniquely determine that Ritika was born in 1954.

Conclusion: Neither statement alone is sufficient, but when combined they provide enough information to answer the question.

The correct answer is: Option D – Both I and II are sufficient.

To balance the given redox reaction, we look at the half reactions first:

Reduction Half-Reaction:

${\mathrm{MnO}}_{-}^4+8H^{+}+5e^{-}⟶{\mathrm{Mn}}^{2+}+4H_{2}O$

Oxidation Half-Reaction:

$H_2{C}_2{O}_4⟶2{\mathrm{CO}}_2+2H^{+}+2e^{-}$

To balance the charges and the number of electrons transferred in both half-reactions, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5:

Reduction Half-Reaction (multiplied by 2):

$2{\mathrm{MnO}}_{-}^4+16H^{+}+10e^{-}⟶2{\mathrm{Mn}}^{2+}+8H_{2}O$

Oxidation Half-Reaction (multiplied by 5):

$5H_2{C}_2{O}_4⟶10{\mathrm{CO}}_2+10H^{+}+10e^{-}$

Adding these two balanced half-reactions gives the overall balanced redox reaction:

$2{\mathrm{MnO}}_{-}^4+5H_2{C}_2{O}_4+6H^{+}⟶2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+8H_{2}O$

From this balanced equation, we can deduce the values of the coefficients:

$𝑥=2$

$𝑦=5$

$𝑚=2$

$𝑛=10$

Thus, the sum $𝑥+𝑦+𝑚+𝑛=2+5+2+10=19$. Therefore, the sum of $𝑥,𝑦,𝑚,$ and $𝑛$ is 19.

To determine the standard enthalpy of formation of ICl(g), we consider the following reaction:

$\frac{1}{2}{I}_2\left(𝑔\right)+\frac{1}{2}{\mathrm{Cl}}_2\left(𝑔\right)⟶\mathrm{ICl}\left(𝑔\right)$

We have been provided with the following data:

Dissociation of Cl${}_2$:

${\mathrm{Cl}}_2\left(𝑔\right)⟶2\mathrm{Cl}\left(𝑔\right),\Delta 𝐻=242.3\text{kJ/mol}$

Dissociation of I${}_2$:

$I_2\left(𝑔\right)⟶2I\left(𝑔\right),\Delta 𝐻=151\text{kJ/mol}$

Dissociation of ICl:

$\mathrm{ICl}\left(𝑔\right)⟶I\left(𝑔\right)+\mathrm{Cl}\left(𝑔\right),\Delta 𝐻=211.3\text{kJ/mol}$

Enthalpy of Sublimation of I${}_2$:

$I_2\left(𝑠\right)⟶I_2\left(𝑔\right),\Delta 𝐻=62.8\text{kJ/mol}$

To calculate the enthalpy change of the given reaction, we use the following steps:

$\text{Required enthalpy change}=\frac{1}{2}\left(\Delta {𝐻}_{{\text{I}}_2}\right)+\frac{1}{2}\left(\Delta {𝐻}_{{\text{Cl}}_2}\right)-\Delta {𝐻}_{\text{ICl}}+\frac{1}{2}\left(\Delta {𝐻}_{\text{Sublimation}}\right)$

Substituting the given values, we have:

$\Delta 𝐻=\frac{1}{2}\left(151\right)+\frac{1}{2}\left(242.3\right)-211.3+\frac{1}{2}\left(62.8\right)$

Calculating this gives:

$\Delta 𝐻=16.8\text{kJ/mol}$

Thus, the standard enthalpy of formation of $\mathrm{ICl}\left(𝑔\right)$ is approximately $16.8\text{kJ/mol}$.

Sucrose $⟶$ No dissociation

$\begin{array}{cc}& \mathrm{NaCl}⟶{\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}\\ & {\mathrm{CaCl}}_2⟶{\mathrm{Ca}}^{2+}+2{\mathrm{Cl}}^{-}\end{array}$

Elevation in boiling point is directly proportional to the number of particles and ${\mathrm{CaCl}}_2$ solution contains more number of particles.

Hence, $\Delta {𝑇}_{𝑏}$ will be maximum for ${\mathrm{CaCl}}_2$ and so will be the boiling point.

Lower the electronegativity, higher is the tendency of the atom/group to donate its pair of electrons and hence higher is nucleophilicity. Now among $F\left(4.0\right),O\left(3.5\right),N\left(3.0\right)$ and $C\left(2.5\right),C$ has the lowest electronegativity and hence, ${\mathrm{CH}}_{-}^3$ has highest nucleophilicity.

Since $3^{\circ }$ alkyl halides prefer to under go elimination rather than substitution, therefore sodium tert-butoxide react with tert-butyl bromide or chlorides to form elimination product isobutylene rather than substitution product di-tert-butyl ether.

The molecules/ions that have the same bond order are:

Explanation

For $O_{2-}^2$ (18 electrons):

Electron Configuration: $𝐾𝐾𝜎\left(2𝑠{)}^2{𝜎}^{\ast }\right(2𝑠{)}^2𝜎\left(2{𝑝}_{𝑧}{)}^2𝜋\right(2{𝑝}_{𝑥}{)}^2{𝜋}^{\ast }\left(2{𝑝}_{𝑦}{)}^2{𝜋}^{\ast }\right(2{𝑝}_{𝑥}{)}^2{𝜋}^{\ast }(2{𝑝}_{𝑦}{)}^2$

Bond Order (B.O.) calculation:

$\text{B.O.}=\frac{\text{Number of bonding electrons}-\text{Number of antibonding electrons}}{2}=\frac{8-6}{2}=1$

For $B_2$ (10 electrons):

Electron Configuration: $𝐾𝐾𝜎\left(2𝑠{)}^2{𝜎}^{\ast }\right(2𝑠{)}^2𝜋\left(2{𝑝}_{𝑥}{)}^1𝜋\right(2{𝑝}_{𝑦}{)}^1$

Bond Order (B.O.) calculation:

$\text{B.O.}=\frac{\text{Number of bonding electrons}-\text{Number of antibonding electrons}}{2}=\frac{4-2}{2}=1$

Thus, $O_{2-}^2$ and $B_2$ both have a bond order of 1, indicating they have identical bond orders.

The correct order (decreasing) of energy of 2 s orbital for given atom is

${𝐸}_{2𝑠}\left(H\right)>{𝐸}_{2𝑠}\left(\mathrm{Li}\right)>{𝐸}_{2𝑠}\left(\mathrm{Na}\right)>{𝐸}_{2𝑠}\left(K\right)$

The reason is that as atomic number (nuclear charge) increases, all the orbitals are pulled closer to the nucleus. Closes in an orbital to the nucleus, less is its energy.

The complete reaction is as follows,

The explanation for calculating the osmotic pressure of a solution formed by dissolving 25 mg of $K_2{\mathrm{SO}}_4$ in 2 liters of water at 25°C is detailed below:

Mass and Volume Information:

Mass of $K_2{\mathrm{SO}}_4$ dissolved = 25 mg = 0.025 g

Volume of the solution = 2 L

Temperature = 25°C = 298 K

Molar Mass Calculation:

The molar mass of $K_2{\mathrm{SO}}_4$ = $2\times 39+32+4\times 16=174g/\mathrm{mol}$

Dissociation in Water:

$K_2{\mathrm{SO}}_4$ dissociates completely as:

$K_2{\mathrm{SO}}_4⟶2K^{+}+{\mathrm{SO}}_{2-}^4$

Number of ions produced per formula unit = 3 (2 $K^{+}$ and 1 ${\mathrm{SO}}_{2-}^4$)

Osmotic Pressure Calculation using Van’t Hoff Factor (i):

Van’t Hoff factor $𝑖=3$

The formula for osmotic pressure is:

$𝜋=𝑖𝐶𝑅𝑇$

Here, $𝐶=\frac{𝑛}{𝑉}$, where $𝑛$ is the moles of solute, and $𝑉$ is the volume in liters.

Calculate moles of $K_2{\mathrm{SO}}_4$:

$𝑛=\frac{0.025g}{174g/\mathrm{mol}}$

Substitute the known values into the osmotic pressure formula:

$𝜋=3\times \frac{0.025}{174}\times \frac{1}{2}\times 0.0821\times 298$

Solving for $𝜋$, we find:

$𝜋=5.27\times {10}^{-3}\mathrm{atm}$

Thus, the osmotic pressure of the solution is $5.27\times {10}^{-3}\mathrm{atm}$.

Esters react with Grignard reagent to form tertiary $\left(3^{\circ }\right)$ alcohols. The complete reaction is given as follows.

The correct match is A-IV, B-I, C-II, D-III

$\bullet$ He has highest ionisation energy due to inert gas configuration.

$\bullet$ Cl has highest electron affinity among given atoms.

$\bullet$ Ca is the most electropositive element among given atoms.

$\bullet$ Li is strongest reducing agent among given atoms.

To find the molality of the solution, we use the formula:

$\text{Molality of solution}=\frac{\text{Mass of solute (in moles)}}{\text{Mass of solvent (in kg)}}$

Given that a 3 M NaCl solution has 3 moles of NaCl in 1 L of solution, we can calculate as follows:

Calculate the mass of the solution:

The density of the solution is 1.25 g/mL.

Volume of the solution = 1 L = 1000 mL.

Mass of 1 L of the solution = Volume × Density = $1000\text{mL}\times 1.25\text{g/mL}=1250\text{g}$.

Calculate the mass of the solute (NaCl):

Molar mass of NaCl = 58.5 g/mol.

Mass of 3 moles of NaCl = $3\times 58.5=175.5\text{g}$.

Determine the mass of the solvent:

Mass of solvent = Total mass of solution - Mass of solute

Mass of solvent = $1250\text{g}-175.5\text{g}=1074.5\text{g}=1.0745\text{kg}$.

Calculate the molality:

$\text{Molality}=\frac{3\text{mol}}{1.0745\text{kg}}=2.7\text{mol/kg}$

Therefore, the molality of the solution is 2.7 mol/kg.

Due to $+𝑅$ effect of Ph group and $+𝐼$ effect of ${\mathrm{CH}}_3$ group, the magnitude of positive charge on carbon atom of $C=O$ group decreases to such an extent that weak nucleophilic like ${\mathrm{NaHSO}}_3$ does not add to the carbonyl group of acetophenone.

Electronegativity difference between $N\left(3.0\right)$ and $\mathrm{Cl}\left(3.0\right)$ is zero and hence, $N-\mathrm{Cl}$ bonds are non-polar. As a result, the overall dipole moment of ${\mathrm{NCl}}_3$ molecule and its direction is just the dipole moment of the lone pair of electrons.

On the other hand, $N-\mathrm{Br}(3.0-2.8)$, $N-I(3.0-2.5)$ and $N-H(3.0-2.1)$ are polar and hence, contribute towards the overall dipole moment of the respective molecules. Since the EN difference is highest in case of $N-H$ bonds, therefore, ${\mathrm{NH}}_3$ has the highest dipole moment.

Number of radial nodes $=𝑛-𝐼-1$

For $3𝑠$, radial nodes $=3-0-1=2$

For $2𝑝$, radial nodes $=2-1-1=0$

Hence, the number of radial nodes are 2, and 0 respectively.

To determine the change in the rate constant when the temperature increases from 298 K to 308 K, we can use the Arrhenius equation.

The Arrhenius equation is given by:

$\log 𝐾=\log 𝐴-\frac{{𝐸}_{𝑎}}{2.303𝑅𝑇}$

From this formula, we can derive the ratio of the rate constants at two different temperatures ${𝑇}_1$ and ${𝑇}_2$:

$\frac{\log {𝐾}_{308}}{\log {𝐾}_{298}}=\frac{{𝐸}_{𝑎}}{2.303𝑅}\left[\frac{{𝑇}_2-{𝑇}_1}{{𝑇}_1\times {𝑇}_2}\right]$

Given that the activation energy, ${𝐸}_{𝑎}$, is $9.0{\text{ kcal mol}}^{-1}$, and substituting the values, we have:

$\frac{\log {𝐾}_{308}}{\log {𝐾}_{298}}=\frac{9.0}{2.303\times 2\times {10}^{-3}}\left[\frac{308-298}{308\times 298}\right]$

Calculating this gives:

$\frac{\log {𝐾}_{308}}{\log {𝐾}_{298}}=0.2129$

Thus, the rate constant at 308 K compared to 298 K is:

${𝐾}_{308}=1.632{𝐾}_{298}$

This means the percentage increase in the rate constant is:

$\%\text{ increase}=\frac{{𝐾}_{308}-{𝐾}_{298}}{{𝐾}_{298}}\times 100\approx 63\%$

This indicates that the rate constant increases by approximately 63% when the temperature is raised from 298 K to 308 K.

1 mole of Al, i.e. 27 g of Al are deposited by charge $=3F$

$\therefore 4.5g$ of Al will be deposited by charge

$=\frac{3}{27}\times 4.5F=0.5F$

2 F of charge liberate $H_2=1\mathrm{mol}=22.4$ litre at STP

$\therefore 0.5F$ of charge will liberate $H_2=\frac{22.4}{2}\times 0.5$

= 5.6 L at STP

On acidification of potassium chromate solution, yellow colour changes to orange due to reformation of dichromate. This suggests that monocentric complex, i.e., potassium monochromate is converted into dicentric complex, i.e. potassium dichromate. The complete reaction is as follows.

$\underset{\begin{array}{c}\text{ Potassium }\\ \text{ chromate }\end{array}}{2K_2{\mathrm{CrO}}_4}+H_2{\mathrm{SO}}_4⟶\underset{\begin{array}{c}\text{ Potassium }\\ \text{ dichromate }\end{array}}{K_2{\mathrm{Cr}}_2{O}_7}+K_2{\mathrm{SO}}_4+H_{2}O$

To determine how long water should be electrolyzed with a current of 100 amperes to release sufficient oxygen to completely burn 27.66 g of diborane, consider the following:

The complete combustion reaction is:

$B_2{H}_6+3O_2⟶B_2{O}_3+3H_{2}O$

Given that 27.66 g of $B_2{H}_6$ corresponds to 1 mole of $B_2{H}_6$, it requires 3 moles of $O_2$ for complete combustion. The electrolysis of water can be represented as:

$H_{2}O⟶H_2+\frac{1}{2}{O}_2$

For $\frac{1}{2}$ mole of $O_2$, 2 Faradays of electricity are required. Therefore, to produce 3 moles of $O_2$, the requirement is:

$3\text{ moles of }{O}_2=3\times 2\times \text{Faraday}=12\text{ Faraday}$

The total charge $𝑄$ (in coulombs) needed is calculated as:

$12\times 96500=𝑖\times 𝑡$

Given that $𝑖=100\text{A}$, substitute into the equation:

$12\times 96500=100\times 𝑡\to 𝑡=\frac{12\times 96500}{100}=3.2\text{hours}$

Thus, the time required for electrolysis is 3.2 hours.

Mixture A: 1‑butyne and 2‑butyne

 • The key difference is that 1‑butyne is a terminal alkyne (it has a hydrogen attached to the sp‐carbon) while 2‑butyne is an internal alkyne.

 • Terminal alkynes can react with ammoniacal silver nitrate (the basis of Tollen’s reagent) to form an insoluble silver acetylide.

 • Thus, adding Tollen’s reagent (II) will distinguish 1‑butyne from 2‑butyne.

Mixture B: Benzene and CH₂=CH₂

 • Benzene is aromatic and quite stable, whereas ethylene (CH₂=CH₂) is an alkene with a reactive double bond.

 • Baeyer’s reagent (a cold alkaline KMnO₄ solution, III) rapidly oxidizes alkenes like ethylene (decolorizing the purple solution) but leaves benzene unaffected.

 • So, mixture B is best distinguished using Baeyer’s reagent (III).

Mixture C: C₆H₆ and C₆H₁₄

 • Here we have benzene (an aromatic compound) and hexane (a saturated alkane).

 • Aromatic compounds tend to burn with a sooty (smoky) flame because of their high carbon content and unsaturation, while alkanes burn with a non‐sooty, luminous flame.

 • Therefore, the “shooty flame test” (IV) can differentiate benzene from hexane.

Mixture D: Benzene and Phenol

 • Both are aromatic, but phenol possesses an –OH group, which greatly increases its boiling point compared to benzene.

 • In the presence of zinc dust, a specific distillation technique (I) can exploit this difference; benzene distills over while phenol (which complexes with zinc) does not.

 • Hence, zinc dust distillation (I) is used to differentiate benzene from phenol.

Matching these results:

 A – Tollen’s reagent (II)

 B – Baeyer’s reagent (III)

 C – Shooty flame test (IV)

 D – Zinc dust distillation (I)

Looking at the options, Option A corresponds to:

 A-II, B-III, C-IV, D-I

Thus, the correct answer is Option A.

Potassium dichromate oxidises ferrous oxalate in acidic medium as follow,

$\begin{array}{cc}& 2{\mathrm{FeC}}_2{O}_4+{\mathrm{Cr}}_2{O}_{2-}^7+14H^{+}\\ & ⟶2{\mathrm{Fe}}^{3+}+2{\mathrm{Cr}}^{3+}+4{\mathrm{CO}}_2+7H_{2}O\end{array}$

Since, 2 moles of ${\mathrm{FeC}}_2{O}_4$ are oxidised by 1 mole of ${\mathrm{Cr}}_2{O}_{2-}^7$ therefore, 1 mole of ${\mathrm{FeC}}_2{O}_4$ will be oxidised by 0.5 mole of ${\mathrm{Cr}}_2{O}_{2-}^7$.

Complex of type ${\mathrm{Ma}}_3{b}_3$ exists in facial (fac) and meridianal (mer) geometrical isomers.

Since, $-𝐼$ effect of -Cl is stronger than its $+𝑅$ effect, therefore, it deactivates the ring towards electrophilic substitution. However, due to $+𝐼$ and hyperconjugation effect of the ${\mathrm{CH}}_3$ group, electron density increases at o-positions. Therefore, electrophilic substitution will be the fastest at position 4 .

Let's first find product A.

The final product B will be

Hence, option (c) is correct.

${𝑑}^0(T^{4+},{\mathrm{Sc}}^{3+})$ and ${𝑑}^{10}\left({\mathrm{Zn}}^{2+}\right)$ species will not absorb visible light and are colourless whereas ${𝑑}^3\left({\mathrm{Cr}}^{3+}\right)$ can undergo $𝑑-𝑑$ transition on absorption of visible light of a particular wavelength and will be coloured. Thus, ${\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}$ is expected to absorb visible light.

Pyridine hydrochloride is a salt formed from the combination of a weak base, pyridine, and a strong acid. When calculating the pH of its solution, you can use the following formula:

$\mathrm{pH}=-\frac{1}{2}[\log {𝐾}_{𝑤}-\log {𝐾}_{𝑏}+\log 𝐶]$

In this context:

$\mathrm{pH}=3.44$

$𝐶=0.02\text{M}$

Setting up the equation with these values:

$3.44=-\frac{1}{2}[-14-\log {𝐾}_{𝑏}+\log (2\times {10}^{-2}\left)\right]$

Solving for $\log {𝐾}_{𝑏}$:

$\log {𝐾}_{𝑏}=-8.82$

Converting this logarithmic form to the ionization constant:

${𝐾}_{𝑏}={10}^{-8.82}\approx 1.5\times {10}^{-9}$

Thus, the ionization constant of pyridine is $1.5\times {10}^{-9}$.

${\mathrm{NO}}_2$ contains one electron on N . It dimerises to give $N_2{O}_4$ where single electron is paired up to form $N-N$ bond in $N_2{O}_4$ and paramagnetic ${\mathrm{NO}}_2$ changes to diamagnetic $N_2{O}_4$.

The intermediate product (L) is

Now at 300 K, (L) is heated in presence of HCl.

Hence, the correct value of M is given in option (a).

Ionisation energy is the energy required to remove an electron from outer shell of an atom. Since, on moving from left to right in a period, size decreases, thus larger amount of energy is required to remove an electron, i.e. ionisation energy increases.

Although the size of O -atom is smaller as compared to N but the ionisation energy of N is higher.

This is because in case of $𝑁$, the electron is to be removed from stable half filled $2{𝑝}^3$ orbital. Hence, the correct order is

$N>O<F$

To determine the solubility product constant, ${𝐾}_{\text{sp}}$, for the reaction:

${\mathrm{Ag}}_2{\mathrm{CO}}_3\left(s\right)\rightleftharpoons 2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CO}}_{2-}^3\left(\mathrm{aq}\right)$

we begin with the understanding that the equilibrium constant expression for the reaction in terms of ${𝐾}_{\text{sp}}$ is:

${𝐾}_{\text{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CO}}_{2-}^3\right]$

Since ${\mathrm{Ag}}_2{\mathrm{CO}}_3$ is a solid, its concentration does not appear in the expression.

To relate this to the Gibbs free energy change, $\Delta {𝐺}^{\circ }=+63.3\mathrm{kJ}/\mathrm{mol}$, we use the following equation:

$\Delta 𝐺=-2.303𝑅𝑇\log 𝐾$

where:

$𝑅$ is the universal gas constant, $8.314J/\mathrm{mol}\cdot K$,

$𝑇$ is the temperature in Kelvin ($298K$ for ${25}^{\circ }C$).

Plugging in the values, we solve for $\log {𝐾}_{\text{sp}}$:

$63300=-2.303\times 8.314\times 298\times \log {𝐾}_{\text{sp}}$

Upon solving the above equation, we determine:

${𝐾}_{\text{sp}}=8\times {10}^{-12}$

The correct answer is Option A: Vitamin B complex.

Here’s why:

Fat Soluble Vitamins: Vitamins A, D, E, and K are fat soluble, meaning they dissolve in fats and can be stored in the body's fatty tissues.

Water Soluble Vitamins: The Vitamin B complex (along with Vitamin C) are water soluble, meaning they dissolve in water and excess amounts are generally excreted by the body.

Since Vitamin B complex does not dissolve in fat, it is not a fat soluble vitamin.

The reaction is:

$𝐴\left(𝑔\right)⟶𝐵\left(𝑔\right)$

This is a first-order reaction.

Initial concentration of reactant A, $[𝐴{]}_0=0.2\mathrm{mol}{L}^{-1}$.

Time elapsed, $𝑡=10\mathrm{minutes}$.

Concentration of product B after 10 minutes, $[𝐵{]}_{𝑡}=0.18\mathrm{mol}{L}^{-1}$.

Our goal is to find the rate constant, $𝑘$.

First, we need to find the concentration of reactant A that is left after 10 minutes, which we call $[𝐴{]}_{𝑡}$.

Look at the stoichiometry of the reaction: $𝐴⟶𝐵$. For every one mole of A that reacts, one mole of B is formed.

So, the concentration of A that has reacted is equal to the concentration of B that has been formed.

Concentration of A reacted = Concentration of B formed = $0.18\mathrm{mol}{L}^{-1}$.

Now, we can calculate the concentration of A remaining after 10 minutes:

$[𝐴{]}_{𝑡}=\text{Initial concentration of A}-\text{Concentration of A reacted}$

$[𝐴{]}_{𝑡}=[𝐴{]}_0-0.18$

$[𝐴{]}_{𝑡}=0.2-0.18=0.02\mathrm{mol}{L}^{-1}$

So, after 10 minutes, the concentration of A is $0.02\mathrm{mol}{L}^{-1}$.

Now, we will use the integrated rate law for a first-order reaction. You might remember this from your NCERT textbook:

$𝑘=\frac{2.303}{𝑡}\log \frac{[𝐴{]}_0}{[𝐴{]}_{𝑡}}$

where:

• $𝑘$ is the rate constant.

• $𝑡$ is the time.

• $[𝐴{]}_0$ is the initial concentration of the reactant.

• $[𝐴{]}_{𝑡}$ is the concentration of the reactant at time $𝑡$.

Let's substitute the values we have into this equation:

• $𝑡=10\min$

• $[𝐴{]}_0=0.2\mathrm{mol}{L}^{-1}$

• $[𝐴{]}_{𝑡}=0.02\mathrm{mol}{L}^{-1}$

$𝑘=\frac{2.303}{10\min }\log \frac{0.2}{0.02}$

Now, let's simplify the expression inside the logarithm:

$\frac{0.2}{0.02}=10$

So, the equation becomes:

$𝑘=\frac{2.303}{10\min }\log \left(10\right)$

We know that the value of $\log \left(10\right)$ (to the base 10) is 1.

$𝑘=\frac{2.303}{10\min }\times 1$

$𝑘=0.2303{\min }^{-1}$

Therefore, the rate constant for the reaction is $0.2303{\min }^{-1}$.

Looking at the options provided, this matches Option A.

So, the correct answer is Option A.

lodoform test is carried out in hot alkaline medium. Under these condition esters get hydrolysed to give the corresponding alcohol. Now ethyl acetate give ethyl alcohol which respond to iodoform test to give yellow ppt. of ${\mathrm{CH}}_3$, while methyl acetate will give methanol which does not give iodoform test.

${\mathrm{CH}}_3{\mathrm{COOC}}_2{H}_5\underset{\Delta }{\overset{I_2/\mathrm{NaOH}}{\to \leftarrow }}$

$[{\mathrm{CH}}_3\mathrm{COONa}+C_2{H}_5\mathrm{OH}]\stackrel{I_2/\mathrm{NaOH}}{\to \leftarrow }{\mathrm{CHI}}_3$

${\mathrm{CH}}_3{\mathrm{COOCH}}_3\underset{\Delta }{\overset{I_2/\mathrm{NaOH}}{\to \leftarrow }}$

$[{\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{CH}}_3\mathrm{OH}]\stackrel{I_2/\mathrm{NaOH}}{\to \leftarrow }\underset{\text{ (No reaction) }}{\mathrm{No}{\mathrm{CHI}}_3}$

During osazone formation (when sugars are reacted with phenylhydrazine) the reaction occurs at ${𝐶}_1$ and ${𝐶}_2$ while rest of the position remains intact. Since glucose and fructose differ from each other only in the arrangement of atom at ${𝐶}_1$ and ${𝐶}_2$, therefore they give the same osazone.

In the quantitative determination of ozone, the process involves reacting ozone with an excess of KI at a specific pH and then titrating the resulting liberated iodine ($I_2$) with an appropriate titrant.

When ozone is treated with excess KI solution buffered to a pH of 9.2 using a borate buffer, iodine ($I_2$) is released quantitatively. This liberated iodine is then titrated against a standard solution of sodium thiosulfate (${\mathrm{Na}}_2{S}_2{O}_3$) with starch used as an indicator. Through this process, the amount of ozone ($O_3$) can be accurately calculated.

The reactions involved are as follows:

$2I^{-}+H_{2}O+O_3\stackrel{\text{pH 9.2}}{\to \leftarrow }2{\mathrm{OH}}^{-}+I_2+O_2$

$2{\mathrm{Na}}_2{S}_2{O}_3+I_2⟶{\mathrm{Na}}_2{S}_4{O}_6+2\mathrm{NaI}$

Here, the thiosulfate reacts with the iodine to form sodium tetrathionate and sodium iodide, thereby allowing for the quantitative analysis of ozone.

The correct answer is:

$\text{would have done}$

Here's why:

The sentence "I ……………. the job if you had paid me enough." is a counterfactual conditional sentence, meaning it refers to a situation that did not actually happen.

In such sentences, the typical structure is:

If-clause: Past perfect (e.g., "if you had paid me enough")

Main clause: "would have" + past participle (e.g., "would have done")

Therefore, the complete sentence reads:

"I would have done the job if you had paid me enough."

This structure correctly reflects a hypothetical situation in the past where the condition was not met.