JEE Main 2023 Chemistry April 13 Shift 1 Question Paper is available here for download. Candidates can download official JEE Main 2023 Chemistry Question Paper PDF with Solution and Answer Key for April 13 Shift 1 using the link below. JEE Main Chemistry Question Paper is divided into two sections, Section A with 20 MCQs and Section B with 10 numerical type questions. Candidates are required to answer all questions from Section A and any 5 questions from section B.
JEE Main 2023 Chemistry Question Paper April 13 Shift 1 PDF
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JEE Main 2023 Chemistry Questions with Solutions
Section – A
Question 1:Given below are two statements:
Statement I: Permutit process is more efficient compared to the synthetic resin method for the softening of water.
Statement II: Synthetic resin method results in the formation of soluble sodium salts.
In the light of the above statements, choose the most appropriate answer from the options given below:
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Which one of the following is most likely a mismatch?
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The energy of an electron in the first Bohr orbit of hydrogen atom is \( -2.18 \times 10^{-18} \) J. Its energy in the third Bohr orbit is ------.
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In the above reaction, left hand side and right hand side rings are named as 'A' and 'B' respectively. They undergo ring expansion. The correct statement for this process is:
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Match The following
Column-A Column-B
(a) Nylon 6 I. Natural Rubber
(b) Vulcanized Rubber II. Cross Linked
(c) cis-1, 4-polyisoprene III. Caprolactam
(d) Polychloroprene IV. Neoprene
Choose the correct answer from options given below:
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What happens when a lyophilic sol is added to a lyophobic sol?
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In the reaction given below
In the following reaction ‘X’ is
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2-Methyl propyl bromide reacts with C\(_2\)H\(_5\)O\(^{-}\) and gives 'A' when reacted with C\(_2\)H\(_5\)OH it gives 'B'. The mechanism followed in these reactions and the products 'A' and 'B' respectively are:
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In the reaction given below
The products formed in the above reaction are:
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CIF\(_5\) at room temperature is:
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The pair of lanthanides in which both elements have high third-ionization energy is:
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The third ionization energy refers to the energy required to remove a third electron from an atom. High third ionization energies generally correspond to elements that, after the removal of two electrons, achieve a stable, noble gas-like electronic configuration.
Step 1: Eu (Europium) has the electron configuration \( [Xe] 4f^7 6s^2 \). When two electrons are removed, it achieves the stable \( [Xe] 4f^7 \) configuration, which is stable due to the half-filled \( 4f \)-orbitals. The third ionization energy for Eu is relatively high due to the stability of this configuration.
Step 2: Yb (Ytterbium) has the electron configuration \( [Xe] 4f^{14} 6s^2 \). Upon removal of two electrons, Yb attains the stable \( [Xe] 4f^{14} \) configuration, which is also stable due to the completely filled \( 4f \)-orbitals. As a result, Yb also has a high third ionization energy.
Step 3: Therefore, both Eu and Yb exhibit high third ionization energies due to the stability of their respective electron configurations after the removal of two electrons.
Thus, the correct pair of lanthanides with high third-ionization energy is Eu and Yb. Quick Tip: For lanthanides, high third ionization energies are observed when elements reach a stable electron configuration after the removal of two electrons, often associated with half-filled or fully-filled \( f \)-orbitals.
The mismatched combinations are
A. Chlorophyll - Co
B. Water hardness - EDTA
C. Photography - \( [Ag(CN)_2]^- \)
D. Wilkinson catalyst - \( [(PPh_3)_3RhCl] \)
E. Chelating ligand - D-Penicillamine
Choose the correct answer from the options given below:
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Which of the following statements are not correct?
A. The electron gain enthalpy of F is more negative than that of Cl.
B. Ionization enthalpy decreases in a group of the periodic table.
C. The electronegativity of an atom depends upon the atoms bonded to it.
D. Al\(_2\)O\(_3\) and NO are examples of amphoteric oxides.
Choose the most appropriate answer from the options given below:
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The radical which mainly causes ozone depletion in the presence of UV radiation is:
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The ozone layer plays a crucial role in protecting life on Earth by absorbing harmful ultraviolet (UV) radiation. The depletion of the ozone layer is primarily caused by certain radicals, which break down ozone molecules when exposed to UV radiation.
Step 1: The main radical responsible for ozone depletion is the chlorine radical (Cl\(^\cdot\)). Chlorine atoms, when released into the atmosphere (for example, from chlorofluorocarbons, or CFCs), can undergo photodissociation due to UV radiation and form chlorine radicals.
Step 2: The chlorine radical (Cl\(^\cdot\)) reacts with ozone (O\(_3\)) molecules, breaking them apart into oxygen molecules (O\(_2\)) and individual oxygen atoms (O\). This reaction contributes significantly to the depletion of the ozone layer.
\[ Cl\cdot + O_3 \rightarrow ClO\cdot + O_2 \] \[ ClO\cdot + O\rightarrow Cl\cdot + O_2 \]
These reactions create a catalytic cycle that destroys ozone molecules, with chlorine atoms being reused in the process, making it a very efficient and persistent process of ozone depletion.
Step 3: Other radicals such as NO\(^\cdot\) and OH\(^\cdot\) can also contribute to ozone depletion, but chlorine radicals (Cl\(^\cdot\)) are considered the most important and effective in causing significant ozone layer destruction, particularly in the stratosphere.
Thus, the correct radical responsible for ozone depletion in the presence of UV radiation is Cl\(^\cdot\). Quick Tip: Chlorine radicals are highly effective at destroying ozone molecules. The primary source of chlorine radicals in the atmosphere is chlorofluorocarbons (CFCs), which are now regulated due to their harmful effects on the ozone layer.
In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one?
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The incorrect statement from the following for borazine is:
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Among the following compounds, the one which shows the highest dipole moment is:
Be(OH)\(_2\) reacts with Sr(OH)\(_2\) to yield an ionic salt. Choose the incorrect option related to this reaction from the following:
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Section – B
Question 21:
Solution of 12 g of non-electrolyte (A) prepared by dissolving it in 1000 mL of water exerts the same osmotic pressure as that of 0.05M glucose solution at the same temperature. The empirical formula of A is CH\(_2\)O. The molecular mass of A is --------- g. (Nearest integer)
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We can use the relation between osmotic pressure and molarity to solve this problem. The formula for osmotic pressure is:
\[ \pi = \frac{nRT}{V} \]
Where:
- \( \pi \) is the osmotic pressure,
- \( n \) is the number of moles of solute,
- \( R \) is the ideal gas constant,
- \( T \) is the temperature in Kelvin,
- \( V \) is the volume of the solution in liters.
Given that the osmotic pressures of the two solutions are the same, we can equate the osmotic pressure of the non-electrolyte solution to that of the glucose solution. The formula for the osmotic pressure of a solution is also related to the molarity of the solution by the formula:
\[ \pi = MRT \]
Where \( M \) is the molarity of the solution.
Since both solutions have the same osmotic pressure:
\[ M_{A} = M_{glucose} \]
For glucose, the molarity \( M_{glucose} \) is given as:
\[ M_{glucose} = 0.05 \, M \]
Let the molecular mass of \( A \) be \( M_A \). The molarity of \( A \) is:
\[ M_A = \frac{moles of A}{volume of solution in liters} \]
The number of moles of \( A \) is given by:
\[ moles of A = \frac{mass of A}{M_A} \]
Given the mass of \( A \) is 12 g and the volume is 1 L (since 1000 mL = 1 L), we can write the molarity of \( A \) as:
\[ M_A = \frac{12}{M_A \times 1} \]
Equating the molarity of \( A \) to that of glucose:
\[ \frac{12}{M_A \times 1} = 0.05 \]
Solving for \( M_A \):
\[ M_A = \frac{12}{0.05} = 240 \, g/mol \]
Thus, the molecular mass of \( A \) is \(\boxed{240}\) g/mol. Quick Tip: To calculate the molecular mass using osmotic pressure, remember that the osmotic pressure is proportional to the molarity of the solution. Equating the osmotic pressures of two solutions with known concentrations allows you to find the molecular mass of an unknown solute.
KMnO\(_4\) is titrated with ferrous ammonium sulphate hexahydrate in the presence of dilute H\(_2\)SO\(_4\). The number of water molecules produced for 2 molecules of KMnO\(_4\) is --------.
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20 mL of calcium hydroxide was consumed when it was reacted with 10 mL of unknown solution of H\(_2\)SO\(_4\). Also, 20 mL standard solution of 0.5 M HCl containing 2 drops of phenolphthalein was titrated with calcium hydroxide. The mixture showed pink colour when the burette displayed the value of 35.5 mL, whereas the burette of H\(_2\)SO\(_4\) showed 25.5 mL initially. The concentration of H\(_2\)SO\(_4\) is --------- M. (Nearest integer)
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t\(_{87.5}\) is the time required for the reaction to undergo 87.5% completion and t\(_{50}\) is the time required for the reaction to undergo 50% completion. The relation between t\(_{87.5}\) and t\(_{50}\) for a first order reaction is -------- t\(_{87.5}\). (Nearest integer)
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For a first-order reaction, the time required for a certain percentage of completion is related to the rate constant of the reaction. The integrated rate law for a first-order reaction is given by:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Where:
- \([A]_0\) is the initial concentration of reactant,
- \([A]\) is the concentration of reactant at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.
The time \( t_{50} \) for 50% completion can be found from the relation:
\[ \ln \left( \frac{[A]_0}{[A]_{50}} \right) = k t_{50} \]
Since the reaction goes to 50% completion, \([A]_{50} = \frac{[A]_0}{2}\). Substituting this into the equation:
\[ \ln \left( \frac{[A]_0}{[A]_0/2} \right) = k t_{50} \]
\[ \ln(2) = k t_{50} \]
\[ t_{50} = \frac{\ln(2)}{k} \]
For 87.5% completion, \([A]_{87.5} = 0.125[A]_0\). Using the integrated rate law:
\[ \ln \left( \frac{[A]_0}{[A]_{87.5}} \right) = k t_{87.5} \]
\[ \ln \left( \frac{[A]_0}{0.125[A]_0} \right) = k t_{87.5} \]
\[ \ln(8) = k t_{87.5} \]
\[ t_{87.5} = \frac{\ln(8)}{k} \]
Now, the relation between \( t_{87.5} \) and \( t_{50} \) is:
\[ \frac{t_{87.5}}{t_{50}} = \frac{\frac{\ln(8)}{k}}{\frac{\ln(2)}{k}} = \frac{\ln(8)}{\ln(2)} = \frac{3\ln(2)}{\ln(2)} = 3 \]
Thus, the value of \( x \) is \( \boxed{3} \). Quick Tip: For a first-order reaction, the time required for different percentages of completion is related to the logarithm of the ratio of the initial and final concentrations. The ratio of times for different completion percentages can be derived using this relationship.
A certain quantity of real gas occupies a volume of 0.15 dm\(^3\) at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300 K (When its compressibility factor is 1.4) is \(\times 10^{-4}\) dm\(^3\). (Nearest integer)
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A metal surface of 100 cm\(^2\) area has to be coated with nickel layer of thickness 0.001 mm. A current of 2 A was passed through a solution of Ni(NO\(_3\))\(_2\) for \( x \) seconds to coat the desired layer. The value of \( x \) is -------------. (Nearest integer)
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25.0 mL of 0.050 M Ba(NO\(_3\))\(_2\) is mixed with 25.0 mL of 0.020 M NaF. \( K_{sp} \) of BaF\(_2\) is \( 0.5 \times 10^{-6} \) at 298 K. The ratio of [Ba\(^{2+}\)] [F\(^{-}\)] and \( K_{sp} \) is ------------. (Nearest integer)
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A\(_2\) + B\(_2\) → 2AB. \(\Delta H_f^0 = -200 \, kJ mol^{-1}\)
A\(_2\) and B\(_2\) are diatomic molecules. If the bond enthalpies of A\(_2\), B\(_2\) and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of A\(_2\) is ------------ kJ mol\(^{-1}\). (Nearest integer)
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An organic compound gives 0.220 g of CO\(_2\) and 0.126 g of H\(_2\)O on complete combustion. If the % of carbon is 24, then the % of hydrogen is ------ \(\times 10^{-1}\). (Nearest integer)
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For the given reaction:
The total number of possible products formed by tertiary carbocation of A is --------.
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JEE Main 2023 Chemistry Paper Analysis April 13 Shift 1
JEE Main 2023 Chemistry Paper Analysis for the exam scheduled on April 13 Shift 1 is available here. Candidates can check subject-wise paper analysis for the exam scheduled on April 13 Shift 1 here along with the topics with the highest weightage.
JEE Main 2023 Chemistry Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Chemistry Marks | 100 marks |
| Total Number of Questions Asked | 20 MCQs + 10 Numerical Type Questions |
| Total Number of Questions to be Answered | 20 MCQs + 5 Numerical Type Questions |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
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