JEE Chemistry - MCQ Practice Questions

According to Coulomb's law, lattice energy is proportional to the product of the ionic charges divided by the distance between ions:

Step 1: Consider the ionic charges.

$\text{MgO}$ consists of ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$ ions, giving a charge product of $|+2|\times |-2|=4$, which is larger than that of singly charged ion pairs.

Step 2: Consider the ionic radii.

The ions ${\text{Mg}}^{2+}$ and ${\text{O}}^{2-}$ have relatively small ionic radii, so the interionic distance $r$ is small. A smaller $r$ leads to a larger lattice energy.

Step 3: Combine both factors.

Since $\text{MgO}$ has both a high charge product and a small interionic distance,

this results in a stronger electrostatic attraction and therefore the highest lattice energy among the given options.

Conclusion:

Vapour pressure depends on the following factors:

Step 1: Effect of Temperature

As temperature increases, more molecules have sufficient kinetic energy to escape the liquid phase, so vapour pressure increases. In general,

Step 2: Effect of Intermolecular Forces

The nature of intermolecular forces in the liquid determines how easily molecules can escape into the vapour phase. Liquids with weaker intermolecular forces have higher vapour pressures, and those with stronger intermolecular forces have lower vapour pressures:

Step 3: Factors That Do NOT Affect Vapour Pressure

Surface area and volume of the liquid do not affect the vapour pressure:

Conclusion

Vapour pressure depends only on:

At the critical point, the compressibility factor is defined as

For a van der Waals gas, substituting the critical constants $P_c=\frac{a}{27b^2}$, $V_c=3b$, and $T_c=\frac{8a}{27Rb}$ gives

For most real gases, the experimentally observed value is approximately

Boiling point elevation depends only on the number of solute particles (expressed as molality $m$), not on their chemical nature. The relationship is given by

where $K_b$ is the ebullioscopic (boiling point elevation) constant of the solvent and $m$ is the molality of the solution. Since $\Delta T_b$ depends solely on $m$ (the number of solute particles per kilogram of solvent) and $K_b$ (a property of the solvent alone), it is completely independent of the chemical nature of the solute.

At constant temperature and pressure, the criterion for spontaneity is

where the Gibbs free energy change is defined as

A process is spontaneous when the enthalpy and entropy contributions combine such that $\Delta H-T\Delta S<0$, meaning the system lowers its Gibbs free energy. The final condition is therefore

Step 1: Identify the Arrhenius two-temperature relation

The two-temperature form of the Arrhenius equation is

where $T_1=25+273=298\text{K}$ and $T_2=75+273=348\text{K}$.

Step 2: Substitute the known values

**Step 3: Solve for $k_2/k_1$

Result:

Step 1: Identify the property type.

A property is extensive if it depends on the amount (or size) of the system, and intensive if it does not.

Step 2: Classify each property.

Step 3: Conclusion.

Entropy $S$ scales with the size of the system — doubling the amount of substance doubles the entropy. The remaining properties (refractive index, surface tension, and density) are independent of the amount of substance, making them intensive.

Step 1: Identify the process

For an isothermal process, the temperature $T$ remains constant throughout the expansion.

Step 2: Apply the work formula

The work done by the gas during isothermal expansion is given by:

where $n$ is the number of moles, $R$ is the universal gas constant, $T$ is the absolute temperature, $V_f$ is the final volume, and $V_i$ is the initial volume.

Step 3: Determine the sign

Since the gas expands, we have $V_f>V_i$, which means:

Therefore, the work done by the gas is positive:

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution:

where $P$ is the partial pressure, $x$ is the mole fraction of the dissolved gas, and $K_h$ is the Henry's law constant.

For ${\text{CO}}_2$ dissolved in water at $25^{\circ }\text{C}$, the Henry's law constant is approximately:

Osmotic Pressure

The osmotic pressure of a solution is given by:

where:

Osmotic pressure is a colligative property, meaning it depends on the number of solute particles present in the solution, not on their chemical identity.

**Step 1: Write the relationship between $\Delta G^{\circ }$ and $E_{\text{cell}}^{\circ }$

The standard Gibbs free energy change is related to the standard cell potential by:

where $n$ is the number of moles of electrons transferred, and $F=96500{\text{C mol}}^{-1}$ is the Faraday constant.

Step 2: Substitute the known values

For $n=2$ (typical two-electron transfer), $F=96500{\text{C mol}}^{-1}$, and $E_{\text{cell}}^{\circ }=0.34\text{V}$:

Result:

**Step 1: Determine $\Delta n$

For the reaction $2{\text{NO}}_2\rightleftharpoons {\text{N}}_2{\text{O}}_4$, the change in moles of gas is:

**Step 2: Apply the $K_p$–$K_c$ relationship**

The general relation is:

Substituting $\Delta n=-1$:

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in pressure, the equilibrium shifts to partially oppose that change.

Step 1: Effect of increasing pressure

Increasing pressure favours the side of the equilibrium with fewer moles of gas. That is, for a general reaction

if $a+b>c+d$, increasing pressure shifts the equilibrium to the right (forward direction).

Step 2: Applying this to the specific reaction

For the reaction

the moles of gas on each side are:

Since $\Delta n_g<0$, increasing pressure favours the forward reaction, producing more ${\text{N}}_2{\text{O}}_4$.

Step 3: Evaluating the statement

The claim that *"increasing pressure always favours the forward reaction"* is not universally correct. It is only correct when the forward reaction proceeds in the direction of fewer moles of gas. A statement made without specifying the reaction, or applied to a reaction where $\Delta n_g=0$ (e.g., ${\text{H}}_2+{\text{I}}_2\rightleftharpoons 2\text{HI}$), would be incorrect, since

Conclusion

The statement that increasing pressure favours the forward reaction is only conditionally true — specifically when $\Delta n_g<0$. Without this qualification, the statement is:

Step 1: Write the relationship between conductivity and molar conductivity

The conductivity $\kappa$ is related to the molar conductivity ${\Lambda }_m$ and the concentration $C$ by:

Step 2: Substitute the given values

With ${\Lambda }_m=426{\text{S⋅cm}}^2\text{/mol}$ and $C=0.01\text{mol/L}=0.01\times 10^{-3}{\text{mol/cm}}^3=10^{-5}{\text{mol/cm}}^3$:

Step 3: State the final result

Since the degree of dissociation is $100\%$, the molar conductivity remains ${\Lambda }_m=426{\text{S⋅cm}}^2\text{/mol}$ unchanged, while the conductivity of the solution is:

Step 1: Identify the cathode and anode

In the Zn–Cu cell, copper undergoes reduction and zinc undergoes oxidation, so:

Step 2: Apply the cell potential formula

Result:

Step 1: Integrated Rate Law Method

Each reaction order corresponds to a specific linear plot. For a first-order reaction, a plot of $\ln [\text{A}]$ versus $t$ gives a straight line with slope $-k$:

For a second-order reaction, a plot of $\frac{1}{[\text{A}]}$ versus $t$ gives a straight line with slope $k$:

The order is identified by whichever plot yields a straight line.

Step 2: Differential Rate Law — Initial Rate Method

The rate is measured at the start of the reaction for different initial concentrations. The general rate law is:

Taking the ratio of two experiments:

Solving for $n$ gives the order with respect to $[\text{A}]$.

Step 3: Half-Life Method

For special cases, the half-life $t_{1/2}$ is used. The relationship between half-life and initial concentration depends on the order:

By observing how $t_{1/2}$ changes with $[\text{A}{]}_0$, the order can be identified.

Summary

The three experimental methods are:

Step 1: Recall the freezing point depression formula

The freezing point depression is given by

where $K_f$ is the cryoscopic constant, $m$ is the molality, and $i$ is the van't Hoff factor.

Step 2: Calculate moles of NaCl

Step 3: Calculate molality

The solvent mass is $100\text{ g}=0.1\text{ kg}$, so

Step 4: Determine the van't Hoff factor

NaCl dissociates completely into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$, giving

Step 5: Calculate the freezing point depression

The closest answer is

Step 1: Write the van't Hoff equation

Step 2: Substitute the known values

Here $K_1=0.5$, $T_1=400\text{K}$, $T_2=500\text{K}$, $\Delta H^{\circ }=-92000{\text{J mol}}^{-1}$, and $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$.

Step 3: Evaluate the temperature term

Step 4: Compute the right-hand side

**Step 5: Solve for $K_2$

Rounding to two significant figures:

Both the larger atomic radius and the greater shielding effect in sodium make the valence electron easier to remove compared to hydrogen's single electron.

Step 1: Effective Nuclear Charge

The effective nuclear charge $Z_{\text{eff}}$ experienced by the valence electron is approximated by Slater's rules:

where $Z$ is the atomic number and $\sigma$ is the shielding constant.

However, despite sodium having a higher $Z_{\text{eff}}$ than hydrogen's $Z_{\text{eff}}=1$ in absolute terms, the valence electron in sodium resides in the $n=3$ shell, which is much farther from the nucleus.

Step 2: Distance and Ionization Energy

The ionization energy scales approximately as:

Since the atomic radius of sodium ($r_{\text{Na}}\approx 186\text{pm}$) is far larger than that of hydrogen ($r_{\text{H}}\approx 53\text{pm}$), the combined effect gives:

Step 3: Conclusion

Therefore, the first ionization energies compare as:

The larger atomic radius and greater electron shielding in sodium reduce the attraction between the nucleus and the $3s$ valence electron, making it significantly easier to remove than hydrogen's $1s$ electron.

Step 1: Write the dissolution equilibrium

For the dissolution of AgCl in pure water:

**Step 2: Set up the expression for $K_{sp}$

Let $s$ be the molar solubility of AgCl. Then $[{\text{Ag}}^{+}]=s$ and $[{\text{Cl}}^{-}]=s$, so:

**Step 3: Solve for $s$

Result: