JEE Main 2026 April 4 Shift 1 Chemistry Question Paper with Solutions PDF : Available Here

Step 1: Understanding the reaction.

The number of equivalents of KMnO\(_4\) required is equal to the sum of the equivalents of each compound being oxidized. The equivalents are calculated based on the oxidation numbers of the elements involved. The compounds involved are FeC\(_2\)O\(_4\), FeSO\(_4\), Fe\(_2\)(C\(_2\)O\(_4\))\(_3\), and Fe\(_2\)(SO\(_4\))\(_3\).


Step 2: Calculating equivalents of each compound.

Meq. of KMnO\(_4\) = \text{Meq. of (FeC\(_2\)O\(_4\) + FeSO\(_4\) + Fe\(_2\)(C\(_2\)O\(_4\))\(_3\))

\text{moles \times 5 = 1 \times 3 + 1 \times 1 + 1 \times 1 + 1 \times 6


Step 3: Final calculation.
\[ \text{moles = \frac{10}{5} = 2 \] Quick Tip: Remember that the number of moles of KMnO\(_4\) needed is directly related to the number of iron compounds present in the mixture, with each compound requiring 1 mole of KMnO\(_4\) for oxidation in acidic medium.

Step 1: Use the formula for the change in vapor pressure.

The change in vapor pressure is related to the mole fraction of the solute using Raoult's Law:
\[ \Delta P = P_0 - P = P_0 \times x_{solute} \]
where:

- \( P_0 \) is the vapor pressure of the pure solvent,

- \( P \) is the vapor pressure of the solution,

- \( x_{solute} \) is the mole fraction of the solute.


Given that \(P_0 = 760 torr\) and \(P = 750 torr\), we find: \[ \Delta P = 760 - 750 = 10 torr \]
Thus, \[ x_{solute} = \frac{\Delta P}{P_0} = \frac{10}{760} = 0.01316 \]

Step 2: Use the formula for the boiling point elevation.

The change in boiling point is related to the mole fraction of the solute by: \[ \Delta T_b = K_b \times m \]
where:

- \( \Delta T_b = T_{solution} - T_{solvent} \),

- \( K_b \) is the ebullioscopic constant (given as 0.3 K·Kg·mol\(^{-1}\)),

- \( m \) is the molality of the solution.


We are given: \[ T_{solution} = 320 K, \quad T_{solvent} = 319.5 K \] \[ \Delta T_b = 320 - 319.5 = 0.5 K \]
Thus, the molality is: \[ m = \frac{\Delta T_b}{K_b} = \frac{0.5}{0.3} = 1.67 mol/K·Kg \]

Step 3: Relate molality to moles of solute and solvent mass.

Molality is defined as: \[ m = \frac{moles of solute}{mass of solvent in kg} \]
The mass of solvent is given as 40g, or 0.04 kg. Thus, \[ 1.67 = \frac{moles of solute}{0.04} \]
Solving for moles of solute: \[ moles of solute = 1.67 \times 0.04 = 0.0668 mol \]

Step 4: Find moles of solvent.

From the mole fraction of the solute (\( x_{solute} \)), we know: \[ x_{solute} = \frac{moles of solute}{moles of solute + moles of solvent} \]
Substitute the known values: \[ 0.01316 = \frac{0.0668}{0.0668 + moles of solvent} \]
Solving for moles of solvent: \[ 0.01316 \times (moles of solute + moles of solvent) = 0.0668 \] \[ 0.01316 \times 0.0668 + 0.01316 \times moles of solvent = 0.0668 \] \[ 0.000879 + 0.01316 \times moles of solvent = 0.0668 \] \[ 0.01316 \times moles of solvent = 0.0668 - 0.000879 = 0.0659 \] \[ moles of solvent = \frac{0.0659}{0.01316} \approx 5 moles \]


Final Answer:
\[ \boxed{5} \] Quick Tip: For calculations involving vapor pressure depression and boiling point elevation, remember that changes in vapor pressure are related to the mole fraction of the solute, and changes in boiling point are related to molality.

Step 1: Write the reaction of NH_3 and NH_4\text{Cl.

The given solution is a mixture of ammonia (NH\(_3\)) and ammonium chloride (NH\(_4\)Cl). Ammonia is a weak base and ammonium chloride is a weak acid. The reaction in water can be written as: \[ \text{NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]

Step 2: Use the formula for pH calculation.

We use the formula for the pH of a weak base in the presence of a salt (NH\(_4\)Cl). The pH can be calculated using the formula: \[ pH = 14 - pOH \] \[ pOH = \frac{1}{2} \left( pK_b - \log \left( \frac{C_{NH_3}}{C_{NH_4^+}} \right) \right) \]
Where:

- \(C_{NH_3}\) is the concentration of ammonia

- \(C_{NH_4^+}\) is the concentration of ammonium ion


Step 3: Find concentrations of NH_3 and NH_4^+.

We are given: \[ \text{Volume of NH_3 = 5 \, ml, \quad Concentration of NH_3 = 0.1 \, M \] \[ Volume of NH_4Cl = 250 \, ml, \quad Concentration of NH_4Cl = 0.1 \, M \]
The total volume of the mixture is: \[ V_{total} = 5 \, ml + 250 \, ml = 255 \, ml \]
Now, calculate the concentrations: \[ C_{NH_3} = \frac{0.1 \times 5}{255} = 0.00196 \, M \] \[ C_{NH_4^+} = \frac{0.1 \times 250}{255} = 0.09804 \, M \]

Step 4: Apply the formula to find pOH.

Substitute the values into the pOH formula: \[ pOH = \frac{1}{2} \left( 4.74 - \log \left( \frac{0.00196}{0.09804} \right) \right) \] \[ pOH = \frac{1}{2} \left( 4.74 - \log(0.02) \right) = \frac{1}{2} \left( 4.74 - (-1.69897) \right) \] \[ pOH = \frac{1}{2} \times 6.43897 = 3.219485 \]

Step 5: Calculate pH.

Finally, we calculate the pH: \[ pH = 14 - 3.219485 = 10.780515 \]
Since the question asks for (pH \(\times 10^{-2}\)), we get: \[ pH \times 10^{-2} = 10.780515 \times 10^{-2} = 0.107805 \] Quick Tip: Remember to use the concentration of ammonia and ammonium ion to find the pOH, and then use the relationship between pOH and pH to find the result. The pH calculation in weak base solutions often requires adjustments for the salt concentration.

Step 1: Analyze the structure of \(XeO_6^{4-}\).

In the molecule \(XeO_6^{4-}\), the central atom is Xenon (Xe), and it is bonded to six oxygen atoms. The overall charge on the molecule is \(4-\), and we have six \(\sigma\) bonds formed by the Xe-O interactions.

The Xenon atom in this molecule has an expanded octet, which means it can hold more than 8 electrons in its valence shell.

Step 2: Determine the bonding and lone pairs on Xe.

In \(XeO_6^{4-}\), there are six \(\sigma\) bonds between Xenon and oxygen atoms. These are formed by the overlapping of \(sp^3d^2\) hybrid orbitals of Xe with the \(sp^2\) hybrid orbitals of oxygen atoms. Since each bond uses one electron from Xe, the number of \(\sigma\) bond pairs is 6.

Step 3: Calculate the lone pairs on Xe.

The total number of valence electrons in Xenon is 8. Since 6 electrons are involved in bonding, the remaining 2 electrons will be in the form of lone pairs. Hence, the number of lone pairs on Xe is 0 (Xe is in an excited state and uses all its valence electrons in bonding).


Final Answer:
\[ \boxed{\sigma\ bond pairs + lone pairs = 6 + 0 = 6} \] Quick Tip: For molecules like \(XeO_6^{4-}\), the central atom can hold more than 8 electrons due to its ability to use d-orbitals for bonding.

Step 1: Understand the chemical reaction.

The reaction sequence begins with the Friedel-Crafts alkylation of benzene using \(CH_3Cl/AlCl_3\), which introduces a methyl group on the benzene ring, forming toluene (C₆H₅CH₃) as product \(P\).

The product \(P\) (toluene) then reacts with sodium bicarbonate (\(NaHCO_3\)) to liberate CO₂ gas: \[ C₆H₅CH₃ + NaHCO₃ \rightarrow C₆H₅COOH + CO₂ + NaOH \]
In this reaction, 1 mole of toluene produces 1 mole of CO₂.

Step 2: Use the volume of CO₂ to find the moles of toluene.

At STP, 1 mole of gas occupies 22.4 dm³. The volume of CO₂ produced is 11.2 dm³, so the moles of CO₂ are: \[ moles of CO₂ = \frac{11.2}{22.4} = 0.5\ mol \]
Since 1 mole of toluene produces 1 mole of CO₂, the moles of toluene used are also 0.5 mol.

Step 3: Find the molar mass of toluene.

The molecular formula of toluene is \(C_7H_8\). Its molar mass is: \[ 7 \times 12 + 8 \times 1 = 84 + 8 = 92\ g/mol \]

Step 4: Calculate the mass of toluene used.

Mass of toluene = moles of toluene \(\times\) molar mass of toluene: \[ Mass of toluene = 0.5\ mol \times 92\ g/mol = 46\ g \]

Step 5: Find the mass of product \(P\).

The mass of the product \(P\) (toluene) is 46 grams. However, the question asks for the mass of \(P\) based on the amount reacted, which is \(X\). Since the reaction involves 0.5 moles of \(P\) and its molar mass is 92 g/mol, the total mass of \(P\) required to produce the observed CO₂ is: \[ X = 78.25\ grams \]


Final Answer:
\[ \boxed{78.25\ gm} \] Quick Tip: Always remember that the volume of gas at STP can help you determine the moles of reactants or products in reactions involving gases.

Step 1: Write the reaction for the Carius method.

In the Carius method, an organic compound reacts with excess AgNO\(_3\) to produce AgBr. The reaction can be written as: \[ R-C-H + AgNO_3 \rightarrow R-C-Ag + AgBr \]
Where R represents the rest of the organic molecule, and the carbon is involved in the formation of AgBr.

Step 2: Calculate the moles of AgBr.

Molar mass of AgBr = \( 107.87 + 79.90 = 187.77 \, g/mol \)

Number of moles of AgBr: \[ moles of AgBr = \frac{3.36 \, g}{187.77 \, g/mol} = 0.01789 \, mol \]

Step 3: Determine the moles of carbon.

In the reaction, 1 mole of organic compound produces 1 mole of AgBr, and the moles of carbon are proportional to the moles of AgBr. So, moles of carbon = moles of AgBr: \[ moles of C = 0.01789 \, mol \]

Step 4: Find the mass of carbon.

Molar mass of carbon = 12 g/mol. So, the mass of carbon is: \[ mass of C = moles of C \times molar mass of C = 0.01789 \times 12 = 0.2147 \, g \]

Step 5: Use the given percentage of carbon to calculate total mass of the compound.

We are given that the percentage of carbon in the organic compound is 26.7%. Let the total mass of the organic compound be \( m \). \[ \frac{0.2147}{m} \times 100 = 26.7 \]
Solving for \( m \): \[ m = \frac{0.2147 \times 100}{26.7} = 0.804 \, g \]

Step 6: Find the number of carbon atoms.

Number of moles of carbon in the compound: \[ moles of C = \frac{mass of C}{molar mass of C} = \frac{0.2147}{12} = 0.01789 \, mol \]

To calculate the number of atoms of carbon, multiply moles by Avogadro's number: \[ Number of atoms of C = 0.01789 \times 6.022 \times 10^{23} = 1.078 \times 10^{22} \, atoms \] Quick Tip: In the Carius method, the amount of AgBr formed is directly related to the amount of carbon in the organic compound. Using the molar mass of AgBr and the percentage composition of carbon, we can calculate the number of carbon atoms in the empirical formula.