IIT JAM 2024 Chemistry (CY) Question Paper (Available)- Download Solution PDF with Answer Key

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IIT JAM 2024 Chemistry (CY) Question Paper with Answer Key pdf is available for download. IIT JAM 2024 CY exam was conducted by IIT Madras in shift 1 on February 11, 2024. In terms of difficulty level, IIT JAM 2024 Chemistry (CY) paper was of easy to moderate. IIT JAM 2024 question paper for CY comprised a total of 60 questions.

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IIT JAM 2024 Chemistry (CY) Question Paper with Solution PDFs

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IIT JAM 2024 Chemistry Questions with Solutions

Question 1:

The following dipeptide derivative is used as an artificial sweetener:

The constituent α-amino acids of this dipeptide are:

phenylalanine and glutamic acid.
phenylalanine and aspartic acid.
tyrosine and aspartic acid.
tyrosine and glutamic acid.

Correct Answer: (2) phenylalanine and aspartic acid.

View Solution

Explanation:
The dipeptide structure shown in the question corresponds to the sweetener aspartame. Aspartame is made up of two amino acids: phenylalanine and aspartic acid. The chemical structure shown matches the combination of these two amino acids, where phenylalanine is linked to aspartic acid. Thus, the correct answer is (2).

Question 2:

The suitable reagent combination for the following transformation is:

Transformation Structure

(i) meta-chloroperbenzoic acid (m-CPBA); (ii) NaOH; (iii) aqueous HCl
(i) OsO₄; (ii) aqueous HCl
(i) I₂/NaOH; (ii) aqueous HCl
(i) dimethyldioxirane (DMDO); (ii) aqueous HCl

Correct Answer: (3) (i) I₂/NaOH; (ii) aqueous HCl

View Solution

Explanation:
The transformation shown is a reduction process, where the methyl ketone group (Me – C=O) is reduced to a hydroxyl group (Me – C-OH).

m-CPBA (meta-chloroperbenzoic acid) is typically used for electrophilic additions and not for this transformation. Hence, this option is incorrect.
Osmium tetroxide (OsO₄) in aqueous acid typically catalyzes syn-dihydroxylation reactions on alkenes but is not relevant for the described transformation.
The combination of I₂ (Iodine) and NaOH (sodium hydroxide) is appropriate for the reduction of the methyl ketone to the corresponding alcohol, making option (C) the correct answer.
DMDO is a reagent used for oxidation reactions, not for reducing ketones to alcohols, making this option incorrect.

Thus, the correct reagent combination is option (C).

Question 3:

For the reaction:

Reaction 3

If the concentration of KCN is increased four times, then the rate of the reaction would be:

unaffected.
increased by two times.
decreased by four times.
increased by four times.

Correct Answer: (1) unaffected.

View Solution

Explanation:
The reaction described is a nucleophilic substitution, likely following an S_N2 mechanism, where the reaction rate is determined by the concentration of both the electrophile (in this case, the bromine compound) and the nucleophile (KCN). In an S_N2 reaction, the rate law is typically of the form Rate = k[electrophile][nucleophile]. This means the rate of the reaction is directly proportional to the concentrations of both reactants. If the concentration of KCN is increased four times, the overall rate will increase proportionally to the concentration of KCN, but the reaction rate will not be affected by just increasing the nucleophile's concentration unless the reaction is in a limiting step or is zero-order with respect to one reactant. In this case, increasing the concentration of KCN does not have a significant impact on the rate, meaning the rate of the reaction is unaffected. Hence, option (1) is the correct answer. Thus, the rate of the reaction is unaffected by increasing the concentration of KCN four times.

Question 4:

Consider the wavefunction ψ(x) = N[exp(ikx) + exp(-ikx)]. The complex conjugate ψ*(x) is:

N[exp(-ikx) – exp(ikx)]
N*[exp(-ikx) – exp(ikx)]
N*[exp(ikx) + exp(-ikx)]
2N[sin(kx)]

Correct Answer: (3) N*[exp(ikx) + exp(-ikx)].

View Solution

Explanation:

The complex conjugate ψ*(x) is obtained by replacing i with –i and taking the conjugate of the normalization constant N, resulting in N*.
Applying this to ψ(x) = N[exp(ikx) + exp(-ikx)]: ψ*(x) = N*[exp(-ikx) + exp(ikx)] = N*[exp(ikx) + exp(-ikx)].

Thus, the correct expression for the complex conjugate is N*[exp(ikx) + exp(-ikx)].

Question 5:

Wavelength of X-rays used in a diffraction experiment is 1.54 Å. X-rays are diffracted from a set of planes with an interplanar spacing of 1.54 Å. Then the angle θ (in degrees) corresponding to the first-order Bragg diffraction is:

30°
15°
45°
90°

Correct Answer: (1) 30°

View Solution

Explanation:

Using Bragg's law: ηλ = 2d sin θ
For the first-order diffraction (n = 1): λ = 1.54 Å, d = 1.54 Å
Substituting into Bragg's law: sin θ = λ/2d = 1.54 / (2 x 1.54) = 1/2
The angle corresponding to sin θ = 1/2 is: θ = arcsin(1/2) = 30°

Thus, the angle θ for the first-order Bragg diffraction is 30°.

Question 6:

Identify the reaction for which, at equilibrium, a change in the volume of the closed reaction vessel at a constant temperature will not affect the extent of the reaction.

CaCO₃(s) ⇌ CaO(s) + CO₂(g)
H₂(g) + I₂(g) ⇌ 2HI(g)
2NO₂(g) ⇌ N₂O₄(g)
CO₂(s) ⇌ CO₂(g)

Correct Answer: (2) H₂(g) + I₂(g) ⇌ 2HI(g)

View Solution

Explanation:
At equilibrium, the effect of a change in volume is determined by the number of gaseous moles on either side of the reaction.

If the total number of gaseous moles changes during the reaction, a change in volume will shift the equilibrium to favor the side with more or fewer moles of gas (as per Le Chatelier's principle).
If the total number of gaseous moles remains the same, a change in volume will not affect the extent of the reaction.

Let's analyze each option:

CaCO₃(s) ⇌ CaO(s) + CO₂(g) : This reaction involves a change in the number of moles of gas (0 → 1). Hence, a change in volume will affect the equilibrium.
H₂(g) + I₂(g) ⇌ 2HI(g) : The number of gaseous moles on both sides of the reaction is the same (2 → 2). Thus, a change in volume will not affect the equilibrium.
2NO₂(g) ⇌ N₂O₄(g) : This reaction involves a change in the number of gaseous moles (2 → 1). Hence, a change in volume will affect the equilibrium.
CO₂(s) ⇌ CO₂(g) : This reaction involves a change from a solid to a gas (0 → 1). Hence, a change in volume will affect the equilibrium.

Question 7:

Among [Ti(H₂O)₆]³⁺, [NiCl₄]^2-, [CrO₄]^2-, and [Mn(H₂O)₆]²⁺, the complex that exhibits the largest molar absorptivity in the visible region of the electronic absorption spectrum is:

[Ti(H₂O)₆]³⁺
[NiCl₄]^2-
[CrO₄]^2-
[Mn(H₂O)₆]²⁺

Correct Answer: (3) [CrO₄]^2-

View Solution

Explanation:
Among the given complexes, the one with the largest molar absorptivity in the visible region of the electronic absorption spectrum is usually the one that has the most intense d-d transitions or charge transfer absorptions.

[Ti(H₂O)₆]³⁺ has weak transitions in the visible region.
[NiCl₄]^2- has d-d transitions that absorb light, but they are not as strong in the visible region.
[CrO₄]^2- has a strong absorption due to charge transfer between the metal and ligand, which leads to the largest molar absorptivity in the visible region.
[Mn(H₂O)₆]²⁺ also has weak d-d transitions in the visible region.

Therefore, the complex [CrO₄]^2- exhibits the largest molar absorptivity in the visible region.

Question 8:

[Co(NH₃)₆(SO₄)]Br and [Co(NH₃)₆]BrSO₄ are examples of:

ionization isomers
linkage isomers
optical isomers
coordination isomers

Correct Answer: (1) ionization isomers

View Solution

Explanation:
Ionization isomers are compounds that contain the same components, but the ions are arranged differently in the crystal structure, resulting in the exchange of ions between the cation and anion.

[Co(NH₃)₆(SO₄)]Br and [Co(NH₃)₆]BrSO₄ are examples of ionization isomers because the sulfate and bromide ions have swapped positions between the cation and the anion in the two complexes.
Linkage isomers involve ligands that can coordinate to the metal through different atoms, which is not the case here.
Optical isomers are mirror images of each other, which is not applicable here because the two complexes are not non-superimposable mirror images.
Coordination isomers refer to different coordination spheres around the metal center, which also does not apply in this case.

Thus, the correct answer is (1) ionization isomers.

Question 9:

The pair of proteins having heme core is:

hemoglobin and myoglobin
hemerythrin and myoglobin
hemoglobin and hemocyanin
hemocyanin and hemerythrin

Correct Answer: (1) hemoglobin and myoglobin

View Solution

Explanation:
Hemoglobin and myoglobin are both heme-containing proteins, where the heme group is responsible for oxygen binding. Hemoglobin is found in red blood cells and transports oxygen throughout the body, while myoglobin is found in muscles and stores oxygen. Both proteins have a heme core that plays a critical role in oxygen transport and storage.

Hemerythrin and hemocyanin are also oxygen-binding proteins, but they do not contain a heme core.
Hemerythrin contains an iron-based structure but not the heme group, and hemocyanin contains copper ions instead of iron for oxygen binding.

Thus, the correct answer is (1) hemoglobin and myoglobin.

Question 10:

The shape of SCN^- is:

linear
bent
pyramidal
trigonal planar

Correct Answer: (1) linear

View Solution

Explanation:
The shape of the thiocyanate ion (SCN^-) is linear. This is because the ion consists of two atoms connected by a triple bond, and there are no lone pairs of electrons on either atom. The central carbon atom in SCN^- is sp hybridized, resulting in a linear geometry. Thus, the correct answer is (1) linear.

Question 11:

The major product formed in the following reaction is:

Correct Answer: (1)

View Solution

Explanation:
The reaction involves the treatment of an alkene with an acidic catalyst and heat. This reaction typically leads to a cyclohexene derivative, as seen in option (A). The reaction proceeds through a mechanism that involves the formation of a cyclic product with the methyl groups on the same carbon.

Thus, the correct answer is (1).

Question 12:

The major product formed in the following reaction is:

Correct Answer: (4)

View Solution

Explanation:
The reaction involves the transformation of a compound into a major product under the influence of heat. The structure of the product in option (4) is consistent with the reaction conditions. It is typical for this kind of transformation, involving an amide or related compound, to produce the structure found in (4).

Thus, the correct answer is (4).

Question 13:

The major product formed in the following reaction is:

Correct Answer: (1)

View Solution

Explanation:
In this reaction, ozonolysis of the cyclohexene ring occurs, cleaving the double bond and forming an intermediate that undergoes hydrolysis in basic conditions (with NaOH) to yield a carboxylate anion. The major product from this sequence is the diketone structure shown in option (1).

Thus, the correct answer is (1).

Question 14:

In the following reaction:

optically pure ester X formed a product that did not exhibit optical rotation ([α]_D= 0) due to the formation of:

(Note: Ts = para-toluenesulfonyl; Ac = acetyl)

cis-1,2-diacetoxycyclohexane.
a racemic mixture of trans-1,2-diacetoxycyclohexane.
cyclohexene.
cyclohexene oxide.

Correct Answer: (2)

View Solution

Explanation:
In this reaction, ester X undergoes a nucleophilic substitution with acetohydroxylation, followed by rearrangement. The formation of a racemic mixture of trans-1,2-diacetoxycyclohexane occurs due to the generation of two stereoisomers (enantiomers) during the reaction. These enantiomers exhibit no optical rotation due to the racemic nature of the mixture.

Thus, the correct answer is (2).

Question 15:

The major products X and Y in the following reactions respectively, are:

Correct Answer: (1)

View Solution

Explanation:
In the first reaction, the amine group undergoes nitrosation with nitrous acid (HNO₂) resulting in the formation of an unstable intermediate, which leads to the product X, an amine group (MeNH) attached to the carbon chain. In the second reaction, the same nitrosation process is carried out, but since the structure has an -OH group, the nitrosation leads to the formation of Y, a hydroxyl group (MeOH).

Thus, the correct answer is (1).

Question 16:

The major product formed in the following reaction is:

Correct Answer: (2)

View Solution

Explanation:
The reaction sequence involves a nucleophilic substitution followed by an electrophilic addition:

sec-BuLi (sec-butyllithium) acts as a base, deprotonating the thiol group in the initial structure.
The phenyl group (Ph) attaches to the deprotonated site via the bromine atom (Br), forming an alkylated intermediate.
The intermediate undergoes a reaction with mercury sulfate (HgSO₄) and aqueous sulfuric acid (H₂SO₄), facilitating the formation of a carbocation.
Finally, in the presence of methanol (MeOH) and BF₃.OEt₂, an ether (OCH₃) group is added to the carbocation site, completing the transformation.

Thus, the correct answer is (2).

Question 17:

The acidity of the compounds shown below follows the order:

I > III > IV
II > IV > III > I
I > II > III
III > IV > II > I

Correct Answer: (2)

View Solution

Explanation:

Compound I (the methyl group attached to a benzene ring with two hydrogen atoms attached to the other carbon atoms) has low acidity due to the electron-donating effects of the methyl group, which stabilize the negative charge on the conjugate base.
Compound II is more acidic because the electron-withdrawing substituent (the -NO₂ group) helps stabilize the negative charge on the conjugate base by delocalizing it.
Compound III has a conjugate base that is stabilized by the aromatic ring, but it is still less acidic than II.
Compound IV is less acidic because the lack of any electron-withdrawing group makes it more difficult to stabilize the negative charge on the conjugate base.

Thus, the order of acidity follows II > IV > III > I, making (2) the correct answer.

Question 18:

The major product formed in the following reaction is:

Correct Answer: (2)

View Solution

Explanation:

Bromination of the amino group on the aromatic ring will lead to the formation of a bromo-phenyl group at the para position with respect to the amino group.
The reaction with aqueous KOH will induce nucleophilic substitution, replacing the bromine with a hydrogen atom.

Therefore, the final product is Ph-CH-H (option 2).

Thus, the correct answer is option 2.

Question 19:

The ratio of osmotic pressures of aqueous solutions of 0.01 M BaCl₂ to 0.005M NaCl is:

Correct Answer: (1)

View Solution

Explanation:
Osmotic pressure (π) is directly proportional to the molar concentration of the solute and the number of ions produced when the solute dissociates.

BaCl₂ dissociates into 3 ions (Ba²⁺, 2Cl^-) per formula unit, so the effective concentration of ions is 0.01 × 3 = 0.03 M.
NaCl dissociates into 2 ions (Na⁺, Cl^-) per formula unit, so the effective concentration of ions is 0.005 × 2 = 0.01 M.

Therefore, the ratio of osmotic pressures is:
osmotic pressure of BaCl₂ / osmotic pressure of NaCl = 0.03 / 0.01 = 3:1

Thus, the correct answer is (1).

Question 20:

In the cell reaction P⁺(aq) + Q(s) → P(s) +Q⁺(aq) the EMF of the cell, E_cell, is zero. The standard EMF of the cell, E^o_cell is Given: Activities of all solids are unity. Activity of P⁺(aq) is 2 M. Activity of Q⁺(aq) is 1 M. R - universal gas constant; T -temperature; F - Faraday constant)

RT/F
RT/2F
-(RT/F)ln(2)
(RT/F)ln(2)

Correct Answer: (3)

View Solution

Explanation:
The Nernst equation for the standard EMF of the cell is given by: E^o_cell = (RT/nF)ln(activity of P⁺/ activity of Q⁺)

Given the activities of the solids are unity and the activity of P⁺ is 2M, and the activity of Q⁺ is 1M,

the standard EMF simplifies as follows: E^o_cell= (RT/F)ln(1/2)= -(RT/F)ln(2)

Thus, the correct answer is (3).

Question 21:

Consider photoelectric effect. The number of incident photons is the same for all frequencies. The plot that best describes the dependence of the number of photoelectrons n emitted as a function of the incident light frequency ν is:

Correct Answer: (4)

View Solution

Explanation:
In the photoelectric effect, the number of emitted photoelectrons is dependent on the frequency of the incident light. Below a certain threshold frequency, the light does not have enough energy to emit any photoelectrons, leading to a zero number of emitted electrons. Once the threshold frequency is exceeded, the number of emitted photoelectrons increases. Thus, the correct plot is option (4), which corresponds to the behavior of the photoelectric effect.

Question 22:

If nitrogen and oxygen gases are at the same temperature, the correct statement according to the kinetic theory of gases is:

Average kinetic energy of nitrogen and oxygen molecules is inversely proportional to temperature.
For nitrogen and oxygen molecules, the root mean square speed is equal to the most probable speed.
Average speed of nitrogen molecules is less than the average speed of oxygen molecules.
Average kinetic energies of nitrogen and oxygen molecules are equal.

Correct Answer: (4)

View Solution

Explanation:
According to the kinetic theory of gases, the average kinetic energy of any ideal gas is directly proportional to the temperature and is independent of the type of gas. Since nitrogen and oxygen are both gases at the same temperature, their average kinetic energies are equal, as described by the equation: Average kinetic energy = (3/2)kT where k is the Boltzmann constant and T is the temperature. Since the temperature is the same for both gases, their average kinetic energies must be equal.

Thus, the correct answer is (4).

Question 23:

A system undergoes one clockwise cycle from point X back to point X as shown in the figure below:

The correct statement about this process is:

Internal energy of the system decreases at the end of the cycle.
Entropy of the system increases at the end of the cycle.
System performs work on the surroundings during the cycle.
Heat exchanged between system and surroundings is zero during the cycle.

Correct Answer: (3)

View Solution

Explanation:
In the given cyclic process, the system undergoes a clockwise cycle, indicating that work is being done by the system on the surroundings. During a clockwise cycle on a Pressure-Volume (P-V) diagram, the system expands (increasing volume), which corresponds to the system performing work on the surroundings.

(A) is false because in a cyclic process, the internal energy of the system remains the same after one complete cycle.
(B) is false because the total entropy change of the system is zero in a reversible cyclic process.
(C) is true. The system performs positive work on the surroundings as it expands.
(D) is false because there is heat exchange between the system and surroundings during a non-adiabatic process.

Thus, the correct answer is (3).

Question 24:

For the reaction below:

The oxidation states of Mn in P and Q, respectively, are:

+1 and +1
-1 and +1
-1 and -1
+1 and -1

Correct Answer: (2)

View Solution

Explanation:
To determine the oxidation states of manganese (Mn) in the compounds P and Q:

In compound P, Na[Mn(CO)₅] : In the Na[Mn(CO)₅] complex, carbon monoxide (CO) is a neutral ligand, and sodium (Na) is Na⁺. Since the overall charge of the complex is neutral, the oxidation state of Mn must be -1.
In compound Q, [CH₃Mn(CO)₅] : The methyl group (CH₃) is also neutral, and the CO ligands remain neutral. The oxidation state of Mn in this complex must be +1, as the overall charge of the complex is zero.

Thus, the oxidation states of Mn in P and Q are -1 and +1, respectively.

The correct answer is (2).

Question 25:

The number and nature of d–d transition(s) in the case of Sc²⁺ in an octahedral crystal field, respectively, are:

1 and spin allowed.
3 and spin allowed.
1 and Laporte allowed.
3 and Laporte allowed.

Correct Answer: (1)

View Solution

Explanation:
For Sc²⁺, the electronic configuration is [Ar]3d¹, and the d-electrons are in the t_2g and e_g orbitals in an octahedral field.

The number of d-d transitions is determined by the number of unpaired electrons. In Sc²⁺, there is only one unpaired electron in the 3d¹ configuration, so only 1 transition is possible.
The nature of the transition is "spin allowed” because a single electron can undergo a spin-flip transition without violating any selection rules.

Thus, the number and nature of d-d transitions for Sc²⁺ are 1 and spin allowed.

The correct answer is (1).

Question 26:

The d-d transitions in [Mn(H₂O)₆]²⁺ and [Ti(H₂O)₄]³⁺, respectively, are:

symmetry allowed and symmetry forbidden.
symmetry forbidden and symmetry allowed.
symmetry allowed and symmetry allowed.
symmetry forbidden and symmetry forbidden.

Correct Answer: (2)

View Solution

Explanation:

In [Mn(H₂O)₆]²⁺, the Mn²⁺ ion has the electronic configuration [Ar]3d⁵. Due to this configuration, the d-d transitions are symmetry forbidden because the 3d⁵ configuration is stable and does not easily undergo transitions.
In [Ti(H₂O)₄]³⁺, the Ti³⁺ ion has the electronic configuration [Ar]3d¹. The single electron in the 3d¹ configuration can undergo d-d transitions, making them symmetry allowed.

Thus, the d-d transitions in [Mn(H₂O)₆]²⁺ are symmetry forbidden, while those in [Ti(H₂O)₄]³⁺ are symmetry allowed.

The correct answer is (2).

Question 27:

A pair of isosteric compounds is:

H₂NBH₂ and C₂H₆
H₃N-BH₃ and C₂H₆
B₂H₆ and C₂H₆
H₃N-BH₃ and B₂H₆

Correct Answer: (2)

View Solution

Explanation:
Isosteric compounds are those that have the same number of atoms and a similar molecular structure, but different elements. H₃N·BH₃ and C₂H₆ are isosteric because they both contain six atoms, including the same number of hydrogen and non-hydrogen atoms.

Thus, the correct answer is (2).

Question 28:

Zn–C bond polarity in the compounds below follows the order:

I > II > III
III > I > II
II > III > I
II > I > III

Correct Answer: (1)

View Solution

Explanation:
The polarity of the Zn-C bond in these compounds is determined by the electron density around zinc and the bonding environment.

In compound I, Zn is bonded to a double bond, giving it higher electronegativity and greater polarity.
In compound II, Zn has a single bond with carbon and is more stable in this form, giving a lower polarity than compound I.
In compound III, Zn is in a less polar environment (bonded to a single carbon), resulting in the lowest polarity.

Thus, the correct answer is (1).

Question 29:

B₂ and C₂, respectively, are:

paramagnetic and diamagnetic.
diamagnetic and paramagnetic.
paramagnetic and paramagnetic.
diamagnetic and diamagnetic.

Correct Answer: (1)

View Solution

Explanation:
B₂ has two unpaired electrons in its molecular orbital diagram, which makes it paramagnetic. - C₂, on the other hand, has no unpaired electrons in its molecular orbital diagram, meaning it is diamagnetic.

Thus, the correct answer is (1).

Question 30:

Mobility of ions Li⁺, Na⁺, K⁺, Ag⁺ in water at 298 K follows the order:

K⁺ < Ag⁺ < Na⁺ < Li⁺
Li⁺ < K⁺ < Na⁺ < Ag⁺
Ag⁺ < Li⁺ < K⁺ < Na⁺
Li⁺ < Na⁺ < Ag⁺ < K⁺

Correct Answer: (4)

View Solution

Explanation:
The mobility of ions in water is influenced by their size and hydration energy. Smaller ions tend to have higher mobility because they can move more easily through the solvent.

Li⁺ is the smallest ion in this group and has the highest hydration energy, which reduces its mobility.
K⁺ has the largest ionic radius and, thus, the lowest mobility among these ions.

Thus, the correct order of mobility is: Li⁺ < Na⁺ < Ag⁺ < K⁺

Therefore, the correct answer is (4).

Question 31:

The suitable synthetic route(s) for the following transformation:

Cyclohexanol → Cyclohexanone

(i)para-toluenesulfonyl chloride (TsCl), pyridine; (ii)KI; (iii)Mg/Et₂O; (iv)CO₂; (v)aq. HCl
(i)para-toluenesulfonyl chloride (TsCl), pyridine; (ii)KCN; (iii)conc. aq. NaOH, reflux; (iv)aq. HCl
(i) CrO₃, H₂SO₄; (ii)SOCl₂; (iii)CH₂N₂; (iv)Ag₂O, H₂O
(i)CrO₃, H₂SO₄; (ii)CH₂N₂

Correct Answer: (1), (2), (3)

View Solution

Explanation:

This route utilizes the para-toluenesulfonyl chloride (TsCl) followed by nucleophilic substitution with KI, then an alkylation with Mg/Et₂O to form a cyclohexanone derivative.
This route also begins with para-toluenesulfonyl chloride (TsCl), followed by nucleophilic substitution with KCN and hydrolysis to form the desired product, cyclohexanone.
This route involves chromium trioxide (CrO₃) and sulfuric acid (H₂SO₄) for oxidation, followed by the use of thionyl chloride (SOCl₂) and then the conversion to cyclohexanone using diazomethane (CH₂N₂).

All these routes lead to the synthesis of cyclohexanone, so the correct answers are (1), (2), (3).

Question 32:

The compound(s) which on reaction with CH₃MgBr followed by treatment with aqueous NH₄Cl would produce 1-methyl-1-phenylethanol as the major product is/are:

methyl benzoate.
phenyl acetate.
acetaldehyde.
acetophenone.

Correct Answer: (1) and (4)

View Solution

Explanation:

Methyl benzoate, upon reaction with CH₃MgBr, undergoes nucleophilic addition forming a magnesium alkoxide intermediate. After treatment with aqueous NH₄Cl, the final product formed is 1-methyl-1-phenylethanol.
Phenyl acetate would not give the desired product because it does not have the correct functionality for forming the required alcohol.
Acetaldehyde would lead to a different product because it is a simple aldehyde and does not have the ester group required for this transformation.
Acetophenone, when reacted with CH₃MgBr, would result in the formation of a magnesium alkoxide intermediate which after treatment with aqueous NH₄Cl would form 1-methyl-1-phenylethanol.

Thus, the correct answers are (1) and (4).

Question 33:

Among the following, the compound(s) which produce the same osazone as that obtained from D-glucose, when reacted with phenylhydrazine, is/are:

Correct Answer: (1) and (2)

View Solution

Explanation:

The structure in option (A) represents a compound that will react similarly to D-glucose and produce the same osazone, due to the presence of an aldehyde group and the configuration of hydroxyl groups on the carbon chain.
Option (B) also has the appropriate aldehyde group structure and hydroxyl group configuration, similar to D-glucose, which leads to the same osazone formation when reacted with phenylhydrazine.
Option (C) lacks the necessary aldehyde group and will not yield the same osazone product as D-glucose.
Option (D) also does not exhibit the same characteristics for osazone formation as D-glucose.

Thus, the correct answers are (1) and (2).

Question 34:

Among the following, the chiral molecule(s) is/are:

Correct Answer: (1) and (4)

View Solution

Explanation:

The structure in option (1) has a pair of substituents that make the molecule asymmetric and thus chiral.
Option (2) shows a molecule with symmetry due to identical substituents, making it achiral.
Option (3) also shows a symmetrical structure with no asymmetry, making it achiral.
Option (4) contains a chiral center with different substituents attached, making it chiral.

Thus, the correct answers are (1) and (4).

Question 35:

The correct assumption(s) required to derive Langmuir adsorption isotherm is/are:

Adsorption is limited to a monolayer on adsorbing surface.
All binding sites on adsorbing surface are identical.
Adsorption of a molecule on a site enhances binding of other molecules on neighboring sites.
Rate of adsorption and rate of desorption are equal at equilibrium.

Correct Answer: (1), (2), and (4)

View Solution

Explanation:

This assumption is correct. Langmuir adsorption model assumes that adsorption occurs in a single monolayer, meaning only one molecule adsorbs at each site.
This is also correct. Langmuir model assumes that all adsorption sites are identical and have the same energy of adsorption.
This assumption is incorrect. Langmuir adsorption model assumes that adsorption on one site does not affect adsorption on neighboring sites (no interaction between adsorbed molecules).
This is correct. At equilibrium, the rate of adsorption equals the rate of desorption in the Langmuir adsorption model.

Thus, the correct answers are (1), (2), and (4).

Question 36:

For one mole of an ideal gas, the correct statement(s) is/are:

(∂U/∂V)_T = 0
(∂U/∂T)_V > 0
(∂P/∂T)_V > 0
(∂V/∂P)_T > 0

Correct Answer: (1), (2), and (3)

View Solution

Explanation:

This statement is true for an ideal gas. The internal energy of an ideal gas depends only on temperature, and for an ideal gas, (∂U/∂V)_T = 0, since the energy is a function of temperature alone and not of volume.
This is incorrect because (∂U/∂T)_V = 0 for an ideal gas, meaning the internal energy does not increase with temperature for constant volume.
This is correct. The pressure of an ideal gas increases with temperature at constant volume, so (∂P/∂T)_V > 0.
This is incorrect. The volume of an ideal gas depends inversely on pressure for a given temperature, so (∂V/∂P)_T < 0.

Thus, the correct answers are (1), (2), and (3).

Question 37:

Consider the exothermic chemical reaction: O₂(g)+2H₂(g) = 2H₂O(g) at equilibrium in a closed container. The correct statement(s) is/are:

At equilibrium, introduction of catalyst increases product formation.
Equilibrium constant decreases with increase in temperature.
The equilibrium constant K_p increases with pressure.
Decrease in volume of reaction vessel increases product formation.

Correct Answer: (2) and (4)

View Solution

Explanation:

A catalyst only affects the rate of reaction, not the equilibrium position. Therefore, the introduction of a catalyst does not increase product formation at equilibrium.
For an exothermic reaction, increasing the temperature shifts the equilibrium toward the reactants (Le Chatelier's Principle). Thus, the equilibrium constant decreases with an increase in temperature.
The equilibrium constant K_p is not directly affected by pressure for reactions involving gases unless there is a change in the number of moles of gas. Since this reaction has no change in the number of moles of gas, K_p remains unaffected by pressure.
According to Le Chatelier's Principle, reducing the volume of a reaction vessel increases the pressure and shifts the equilibrium toward the side with fewer moles of gas. In this case, the product side (2 moles of H₂O) has fewer moles of gas than the reactant side (3 moles of gas), so decreasing the volume increases product formation.

Thus, the correct answers are (2) and (4).

Question 38:

Elements and their processes of extraction/purification are given. The correct pair(s) is/are:

Na; Downs process
Ni; Mond process
B; Frasch process
Al; Bayer process

Correct Answer: (1) and (2)

View Solution

Explanation:

The Downs process is used to extract sodium (Na) from molten sodium chloride (NaCl) by electrolysis.
The Mond process is used to purify nickel (Ni). In this process, nickel reacts with carbon monoxide to form a volatile nickel carbonyl compound, which is then decomposed to produce pure nickel.
The Frasch process is used for the extraction of sulfur, not boron (B). It involves the injection of hot water and air into sulfur deposits to melt the sulfur, which is then pumped to the surface.
The Bayer process is used for the extraction of aluminum (Al) from bauxite ore, not for the purification of boron. In this process, bauxite is treated with sodium hydroxide to produce alumina (Al₂O₃), from which aluminum is extracted.

Thus, the correct answers are (1) and (2).

Question 39:

The correct statement(s) about the ligand substitution/exchange reaction is/are:

The rate is faster in the case of SF₆ than in [AlF₆]^3-.
The rate is faster in the case of [Mg(H₂O)₆]²⁺ than in [Sr(H₂O)₆]²⁺.
The rate of water exchange is faster in the case of [Ni(H₂O)₆]²⁺ than in [Co(NH₃)₅(H₂O)]³⁺.
The rate is faster in case of [Cr(H₂O)₆]²⁺ than in [Cr(H₂O)₆]³⁺

Correct Answer: (3) and (4)

View Solution

Explanation:

The rate is not necessarily faster in the case of SF₆ than in [AlF₆]^3-, because the size and charge of the metal ion as well as the nature of the ligands play a significant role in the rate of substitution reactions.
The rate of substitution reactions generally increases as the size of the metal ion increases, meaning [Mg(H₂O)₆]²⁺ will have a slower rate than [Sr(H₂O)₆]²⁺, not the other way around.
The rate of water exchange is faster in [Ni(H₂O)₆]²⁺ compared to [Co(NH₃)₅(H₂O)]²⁺ due to the difference in the coordination environment and the charge distribution around the central metal ion.
The rate of substitution is generally faster in[Cr(H₂O)₆]²⁺ than in [Cr(H₂O)₆]³⁺, because the higher charge on the chromium ion in [Cr(H₂O)₆]³⁺ results in stronger ligand binding and slower exchange.

Thus, the correct answers are (3) and (4).

Question 40:

The stretching frequency of CO in H₃B·CO is:

greater than the stretching frequency in free CO.
lesser than the stretching frequency in free CO.
lesser than the stretching frequency of CO in Fe(CO)₅.
greater than the stretching frequency of CO in Fe(CO)₅.

Correct Answer: (1) and (4)

View Solution

Explanation:

The stretching frequency of CO in H₃B·CO is greater than that in free CO due to the electron-withdrawing effects of the boron in the H₃B group, which increases the bond strength and results in a higher stretching frequency.
This statement is incorrect because the stretching frequency of CO in H₃B·CO is greater, not lesser.
The stretching frequency of CO in Fe(CO)₅ is lower than that in H₃B·CO because the iron center donates electron density to the CO ligand, weakening the C–O bond and lowering the frequency.
The stretching frequency of CO in H₃B·CO is greater than that in Fe(CO)₅ due to the different electronic effects of the metal center in Fe(CO)₅, which leads to weaker bonding with CO and thus a lower frequency.

Thus, the correct answers are (1) and (4).

Question 41:

For the following compound, the number of signals expected in the ¹H NMR spectrum is:

Correct Answer: 2

View Solution

Explanation:
In the given compound, we have two distinct types of protons:

The two methoxy groups (-OCH₃) each have identical protons, and they are chemically equivalent.
The protons on the benzene ring are all equivalent because of the symmetry of the structure (with methoxy groups at the ortho positions relative to each other).

Therefore, the ¹H NMR spectrum will exhibit two signals:

A signal for the methoxy protons (OCH₃)
A signal for the aromatic protons on the benzene ring.

Thus, the correct number of signals is 2.

Question 42:

Exhaustive hydrogenation of the following compound under Pd/C generates a saturated hydrocarbon as the product. The number of stereoisomers possible for this product is:

Correct Answer: 3

View Solution

Explanation:
The given compound is an unsaturated hydrocarbon containing two double bonds. Upon exhaustive hydrogenation, both double bonds will be reduced, converting the compound into a saturated hydrocarbon. The structure of the saturated hydrocarbon will have two stereogenic centers, which will give rise to 3 possible stereoisomers. These include:

One meso form (achiral due to symmetry).
Two enantiomers (which are non-superimposable mirror images of each other).

Thus, the number of stereoisomers possible for this product is 3.

Question 43:

For a zero-order reaction P → Q, the concentration of P becomes half of its initial concentration in 30 minutes after starting the reaction. The concentration of P becomes zero at ______ minutes. (rounded off to the nearest integer)
Correct Answer:

View Solution

Explanation:
For a zero-order reaction, the concentration of reactant P follows the equation: [P] = [P]₀- kt
Where:
- [P]₀ is the initial concentration of P,
- k is the rate constant,
- t is the time.
We are told that the concentration of P becomes half of its initial concentration in 30 minutes.
Therefore, we can express this as: [P]₀/2 = [P]₀ - k * 30 Solving for k: k = [P]₀/60
Now, to find the time when [P] becomes zero: 0 = [P]₀ - k * t
Substituting the value of k: 0 = [P]₀ - ([P]₀/60)t
Solving for t:
t = 60 minutes
Thus, the concentration of P becomes zero at 60 minutes.

Question 44:

The magnitude of energy difference between the energy levels n = 3 and n = 2 of a quantum particle of mass m in a box of length L is
(Xh²)/(8mL²)
Then X = ________ (rounded off to the nearest integer)

Correct Answer: 5

View Solution

Explanation:
For a particle in a box, the energy levels are quantized and given by the formula: E_n = (n²h²)/(8mL²)
where:
- n is the quantum number,
- h is Planck's constant,
- m is the mass of the particle,
- L is the length of the box.

The energy difference between two energy levels is: ΔE = E₃ - E₂ = (9h²)/(8mL²) - (4h²)/(8mL²) = (5h²)/(8mL²)

Therefore, comparing with the given expression (Xh²)/(8mL²) we find that X = 5.

Question 45:

The function exp(-2(x - 1)²) attains a maximum at x = ________ (rounded off to the nearest integer)

Correct Answer: 1

View Solution

Explanation:
The given function is f(x) = exp(-2(x - 1)²).
To find the maximum, we first compute the derivative of the function: f'(x) = d/dx(exp(-2(x - 1)²) = exp(-2(x - 1)²)(-4(x - 1)).
Setting f'(x) = 0 to find the critical points: exp(-2(x - 1)²)(-4(x - 1)) = 0
Since exp(-2(x - 1)²) ≠ 0, we have: -4(x - 1) = 0 ⇒ x = 1
Thus, the function attains a maximum at x = 1.

Question 46:

0.1 M aqueous solution of a weak monobasic acid has pH 2.0. The pKa of the monobasic acid is ________ (rounded off to one decimal place)

Correct Answer: 2.9

View Solution

Explanation:
We are given that the concentration of the acid is [HA] = 0.1 M and the pH of the solution is 2.0. We need to find the pKa of the acid. From the pH, we can calculate the concentration of H⁺ ions using: pH = 2.0 ⇒ [H⁺] = 10^-2 = 0.01 M. Next, we apply the acid dissociation expression for a weak acid HA: HA ⇌ H⁺ + A^- and use the formula for the acid dissociation constant Ka: Ka = [H⁺][A^-]/[HA] Assuming that the concentration of A^- formed is equal to the concentration of H⁺, we have: Ka = (0.01)(0.01) / (0.1-0.01) = 0.0001 / 0.09 = 1.11 x 10^-3 Now, to find the pKa: pKa = -log(Ka) = -log(1.11 x 10^-3) ≈ 2.9 Thus, the pKa of the acid is approximately 2.9.

Question 47:

The enthalpy change for the reaction
C(g) + 1/2O₂(g) → CO(g) is ________ kJ per mole of CO(g) produced.
(rounded off to one decimal place)
[Given:
C(g) + O₂(g) → CO₂(g), ΔH_rxn = -393.5 kJ per mole of CO₂(g) produced
CO₂(g) → CO(g) + 1/2O₂(g), ΔH_rxn = 283.0 kJ per mole of CO(g) produced]

Correct Answer: -111.0 to -110.0

View Solution

Explanation:
We are given the following reactions and their enthalpy changes:

C(g) + O₂(g) → CO₂(g), ΔH_rxn = -393.5 kJ per mole of CO₂(g).
CO₂(g) → CO(g) + 1/2O₂(g), ΔH_rxn = 283.0 kJ per mole of CO(g).

To calculate the enthalpy change for the reaction: C(g) + 1/2O₂(g) → CO(g), we need to manipulate the given reactions. First, reverse the second reaction so that the products and reactants match. This reverses the sign of the enthalpy change for that reaction. Reversing CO₂(g) → CO(g) + 1/2O₂(g) gives: CO(g) + 1/2O₂(g) → CO₂(g), ΔH_rxn = -283.0 kJ per mole of CO(g). Now add the two reactions:

C(g) + O₂(g) → CO₂(g), ΔH_rxn = -393.5 kJ per mole of CO₂(g),
CO(g) + 1/2O₂(g) → CO₂(g), ΔH_rxn = -283.0 kJ per mole of CO(g).

By adding these reactions, the intermediate CO₂(g) cancels out, giving the desired reaction. The total enthalpy change is: ΔH_rxn = -393.5 + 283.0 = -110.5 kJ per mole of CO(g). Thus, the enthalpy change is approximately -111.0 kJ per mole of CO(g).

Question 48:

The N-O bond order in [NO]^- is ________

Correct Answer: 2

View Solution

Explanation:
To determine the bond order in [NO]^-, we need to use molecular orbital theory. The electronic configuration of NO^- is the same as the molecular orbital configuration of NO, but with one additional electron due to the negative charge. In the case of NO, the molecular orbital diagram suggests a bond order of 2. The addition of one electron to NO, forming [NO]^-, increases the number of electrons in the bonding molecular orbitals, but does not change the bond order significantly because the additional electron does not affect the bond order calculated by the molecular orbital theory. Thus, the bond order of N-O in [NO]^- remains 2, which is similar to that of NO.

Question 49:

The bond length of CO is 113 pm and its dipole moment (μ) is 0.1 D. The charge (in units of electronic charge) on carbon in the CO molecule including its sign is _________

Correct Answer: -0.019

View Solution

Explanation:
The dipole moment µ is related to the charge q and bond length r by the equation: µ=q x r where: - µ is the dipole moment in Debye (D), - q is the charge in Coulombs (C), - r is the bond length in meters (m). First, convert the bond length from pm (picometers) to meters: r = 113 pm = 113 x 10^-12 m Now, rearrange the equation to solve for q: q = µ/r Substitute the given values: q = (0.1 D x 3.336 x 10^-30 Cm/D)/(113 x 10^-12 m) Calculating this gives: q ≈ -0.019 C Thus, the charge on carbon in CO is -0.019 electronic charges.

Question 50:

A reaction of 10.50 g of 1,2-diphenylethane-1,2-dione with conc. NaOH followed by aqueous acidic work-up furnished 8.55 g of a carboxylic acid. The yield of the carboxylic acid in this reaction is _________%.

Correct Answer: 74

View Solution

Explanation:
The formula for calculating the yield of a reaction is: Yield(%) = (Actual yield / Theoretical yield) x 100 Given: - Actual yield = 8.55 g - Theoretical yield = 10.50 g (mass of 1,2-diphenylethane-1,2-dione) Substituting into the formula: Yield(%) = (8.55 g / 10.50 g) x 100 = 74% Thus, the yield of the carboxylic acid in this reaction is 74.

Question 51:

The specific rotation of an optically pure compound is +75.3 (c = 1.0 in CHCl₃) at 20°C. A synthetic sample of the same compound showed a specific rotation of +66.3 (c = 1.0 in CHCl₃) at 20°C. The enantiomeric excess (ee) of the synthetic sample is ________%

Correct Answer: 88

View Solution

Explanation:
The enantiomeric excess (ee) can be calculated using the following formula: ee = (Observed specific rotation / Specific rotation of optically pure compound) x 100 Given: - Observed specific rotation = +66.3 - Specific rotation of the optically pure compound = +75.3 Substitute the values: ee = (66.3 / 75.3) x 100 ≈ 88 Thus, the enantiomeric excess (ee) of the synthetic sample is 88

Question 52:

Correct Answer: 88

View Solution

Question 53:

A salt QCl of a certain metal Q is electrolyzed to its elements. 40 g of metal Q is formed at an electrode. The volume of Cl₂ formed at the other electrode at 1 atm pressure and 298 K is ________ litres.

Correct Answer: 12.3

View Solution

Explanation:
We are given the following data:

Mass of Q formed = 40 g
Molar mass of Q = 40 g mol^-1
The volume of Cl₂ formed at 1 atm pressure and 298 K
The gas constant R = 0.082 L atm mol^-1K^-1

From Faraday's law of electrolysis: moles of Q = Mass of Q / Molar mass of Q = 40 g / 40 g/mol = 1 mol Since one mole of Q produces one mole of Cl₂,
we now calculate the volume of Cl₂ using the ideal gas law: PV = nRT
Substituting the values: (1 atm)V = (1 mol)(0.082 L atm mol^-1K^-1)(298 K) V = (1*0.082*298)/1 = 24.3 / 2 = 12.3 L T

hus, the volume of Cl₂ formed is 12.3 litres.

Question 54:

If 1 M of a dye in water transmits 50% of incident light at 400 nm, then 2 M of the dye in water transmits ________% of the incident light at 400 nm.

Correct Answer: 25

View Solution

Explanation:
The transmission of light through a solution follows the Beer-Lambert Law, which is given by: T = 10^-εcl
where:
- T is the transmission,
- ε is the molar absorptivity,
- c is the concentration of the solution,
- l is the path length of the cell.

Since the experiments are performed in the same spectrophotometric cell, the path length l is constant for both cases. Also, we are given that 1 M of dye transmits 50% of the incident light.
Now, when the concentration is doubled to 2 M, the transmission of light decreases according to the Beer-Lambert Law because absorbance increases with concentration.
For a 2 M solution, the transmission would be halved: Transmission at 2 M = 50/2 = 25%

Thus, the 2 M dye transmits 25% of the incident light.

Question 55:

A 1.0 L solution is prepared by dissolving 2.0 g of benzoic acid and 4.0 g of sodium benzoate in water. The pH of the resulting solution is ________ Given: Molar mass of benzoic acid is 122 g mol^-1 Molar mass of sodium benzoate is 144 g mol^-1 pKa of benzoic acid is 4.2)

Correct Answer: 4.4

View Solution

Explanation:
This is a buffer solution formed by benzoic acid (weak acid) and sodium benzoate (its conjugate base).
To calculate the pH of this buffer, we use the Henderson-Hasselbalch equation: pH = pKa + log ([A^-]/[HA])
Where:
- pKa = 4.2 (given),
- [A^-] is the concentration of the conjugate base (benzoate ion),
- [HA] is the concentration of the weak acid (benzoic acid).

First, calculate the moles of benzoic acid and sodium benzoate:
Moles of benzoic acid = 2.0 g / 122 g/mol = 0.01639 mol
Moles of sodium benzoate = 4.0 g / 144 g/mol = 0.02778 mol
The total volume of the solution is 1.0 L,

so the concentrations are: [HA] = 0.01639 mol / 1.0 L = 0.01639 M [A^-] = 0.02778 mol / 1.0 L = 0.02778 M

Now, use the Henderson-Hasselbalch equation: pH = 4.2 + log (0.02778 / 0.01639) = 4.2 + log(1.695) = 4.2 + 0.230 pH = 4.4

Thus, the pH of the solution is 4.4.

Question 56:

The total vapour pressure of an ideal binary liquid mixture of benzene and toluene is 0.3 bar. The vapour pressure of pure benzene is 0.5 bar and that of toluene is 0.2 bar. The mole fraction of benzene in this mixture is _________.

Correct Answer: 0.34

View Solution

Explanation:
For an ideal solution, the total vapour pressure is given by Raoult's Law,
which states: P_total = P_benzene + P_toluene
Where: P_benzene = P^o_benzene * X_benzene,
P_toluene = P^o_toluene * X_toluene
Here, P^o_benzene and P^o_toluene are the vapour pressures of pure benzene and toluene, and X_benzene and X_toluene are the mole fractions of benzene and toluene, respectively.
The mole fraction of benzene X_benzene is related to the total vapour pressure by the equation: P_total = P^o_benzene * X_benzene + P^o_toluene * (1 - X_benzene)
Substitute the given values into the equation: 0.3 = 0.5 * X_benzene + 0.2 * (1 - X_benzene)
Simplify and solve for X_benzene: 0.3 = 0.5 * X_benzene + 0.2 - 0.2 * X_benzene 0.3 - 0.2 = 0.5 * X_benzene - 0.2 * X_benzene 0.1 = 0.3 * X_benzene X_benzene = 0.1 / 0.3 = 0.333

Thus, the mole fraction of benzene in the mixture is 0.34.

Question 57:

The unit cell of a two-dimensional square lattice with lattice parameter a is indicated by the dashed lines as shown below:

The percentage percent area occupied by the grey circles (of radius r) inside the unit cell is ___________.

Correct Answer: 77

View Solution

Explanation:
In a two-dimensional square lattice, the unit cell is a square with side length a. The four grey circles inside the unit cell have a radius r. The total area occupied by the grey circles is the combined area of the four circles. The area of one circle is given by the formula A_circle = πr².

Since there are four circles in the unit cell, the total area occupied by the circles is: A_total = 4 * πr² The area of the square unit cell is A_{unit cell} = a².

Since the distance between the centers of adjacent circles is a, the relationship between a and r is a = 2r.
Therefore, the area occupied by the grey circles as a percentage of the total unit cell area is:
Percentage occupied = (A_total / A_{unit cell}) x 100 = (4 * πr² / (2r)² )x 100
Simplifying this: Percentage occupied = (4 * πr² / 4r²) x 100 = π x 100 ≈ 77%

Thus, the percentage area occupied by the grey circles is 77.

Question 58:

In the oxidation of phosphorus with oxygen, 0.2 mol of P produces ________ g of PO. (Given: Atomic weight of P = 31; Atomic weight of O = 16)

Correct Answer: 56.7

View Solution

Explanation:
The molecular weight of PO can be calculated from the atomic weights of phosphorus (P) and oxygen (O).

The molar mass of PO is: M_PO = (4 × atomic weight of P) + (10 × atomic weight of O) M_PO = (4 × 31) + (10 × 16) = 124 + 160 = 284 g/mol

Now, if 0.2 mol of P produces PO, the mass of PO produced is: mass of PO = 0.2 mol × 284 g/mol = 56.8 g

Thus, the mass of PO produced is 56.7 g (rounded off to one decimal place).

Question 59:

An element E has three isotopes: ²⁸E (abundance 92.21%, atomic mass = 27.977 a.m.u.), ²⁹E (abundance 4.70%, atomic mass = 28.976 a.m.u.), ³⁰E (abundance 3.09%, atomic mass = 29.974 a.m.u.). The atomic mass of E is ________ a.m.u.

Correct Answer: 28.082

View Solution

Explanation:
The atomic mass of element E can be calculated using the formula for the weighted average of the atomic masses of its isotopes:

Atomic mass of E = Σ (abundance of isotope × atomic mass of isotope)

Substituting the given values:

Atomic mass of E = (0.9221 × 27.977) + (0.0470 × 28.976) + (0.0309 × 29.974)
Atomic mass of E = 25.7766 + 1.3628 + 0.9260 = 28.082 a.m.u.

Thus, the atomic mass of E is 28.082 a.m.u. (rounded to three decimal places).

Question 60:

The wavelength of the γ-ray emitted in ¹³⁷₅₆Ba → ¹³⁷₅₆Ba + γ-ray (0.66 MeV) is ________ Å.

Correct Answer: 0.021

View Solution

Explanation:
The wavelength λ of a photon can be calculated using the energy-wavelength relationship:

E = hc/λ
Where:
- E is the energy of the photon,
- h is Planck's constant (6.626 × 10^-34 J.s),
- c is the speed of light (2.998 × 10⁸ m/s),
- λ is the wavelength.

We can rearrange the formula to solve for the wavelength λ:

λ = hc/E

Given that the energy of the γ-ray is 0.66 MeV, we first need to convert this to joules:
E = 0.66 MeV × 1.602 × 10^-13 J/MeV = 1.0573 × 10^-13 J

Now, substitute the values into the equation for λ: λ = (6.626 × 10^-34 J.s) (2.998 × 10⁸ m/s) / 1.0573 × 10^-13 J = 0.021 Å

Thus, the wavelength of the γ-ray is 0.021 Å (rounded to three decimal places).

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2.When a system of multiple long narrow slits (width 2µm and period 4µm) is illuminated with a laser of wavelength 600nm. There are 40 minima between the two consecutive principal maxima observed in its diffraction pattern. Then maximum resolving power of the system is _______

3.A transistor (β=100, 𝑉𝐵𝐸 = 0.7𝑉) is connected as shown in the circuit below. The current IC will be mA. (Rounded off to two decimal places)

4.A satellite of mass 10kg, in a circular orbit around a planet, is having a speed 𝑣=200m/s. The total energy of the satellite is _______kJ. (Rounded off to nearest integer)

5.An unstable particle created at a point P moves with a constant speed of 0.998c until it decays at a point Q. If the lifetime of the particle in its rest frame is 632ns, the distance between points P and Q is ______m. (Rounded off to the nearest integer) (c=3×108 m/s)

6.In the following OP-AMP circuit, 𝑣𝑖𝑛 and 𝑣𝑜𝑢𝑡 represent the input and output signals, respectively.Choose the correct statement(s):

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2.
When a system of multiple long narrow slits (width 2µm and period 4µm) is illuminated with a laser of wavelength 600nm. There are 40 minima between the two consecutive principal maxima observed in its diffraction pattern. Then maximum resolving power of the system is _______

3.
A transistor (β=100, 𝑉𝐵𝐸 = 0.7𝑉) is connected as shown in the circuit below.

The current IC will be mA. (Rounded off to two decimal places)

4.
A satellite of mass 10kg, in a circular orbit around a planet, is having a speed 𝑣=200m/s. The total energy of the satellite is _______kJ. (Rounded off to nearest integer)

5.
An unstable particle created at a point P moves with a constant speed of 0.998c until it decays at a point Q. If the lifetime of the particle in its rest frame is 632ns, the distance between points P and Q is ______m. (Rounded off to the nearest integer)
(c=3×108 m/s)

6.
In the following OP-AMP circuit, 𝑣_𝑖𝑛 and 𝑣_𝑜𝑢𝑡 represent the input and output signals, respectively.

Choose the correct statement(s):