JEE Chemistry — MCQ Practice Questions & Answers

Question 1

Which of the following has the highest lattice energy?

Accepted Answer

Correct answer: B. MgO — According to Coulomb's law, lattice energy is proportional to the product of the ionic charges divided by the distance between ions: U \propto \frac{|q_+|\,|q_-|}{r} **Step 1: Consider the ionic charges.** 	ext{MgO} consists of 	ext{Mg}^{2+} and 	ext{O}^{2-} ions, giving a charge product of |{+2}|	imes|{-2}| = 4 , which is larger than that of singly charged ion pairs. **Step 2: Consider the ionic radii.** The ions 	ext{Mg}^{2+} and 	ext{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 	ext{MgO} has both a high charge product and a small interionic distance, U_{	ext{MgO}} \propto \frac{|{+2}|	imes|{-2}|}{r_{	ext{small}}} this results in a stronger electrostatic attraction and therefore the highest lattice energy among the given options. **Conclusion:** \boxed{	ext{MgO has the highest lattice energy.}}

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A. NaCl

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C. KCl

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D. CaO

Question 2

The vapour pressure of a liquid depends on:

Accepted Answer

Correct answer: B. Temperature and nature of the liquid — 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, 	ext{Vapour pressure} \uparrow \quad 	ext{as} \quad T \uparrow **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: 	ext{Stronger intermolecular forces} \implies 	ext{Lower vapour pressure} **Step 3: Factors That Do NOT Affect Vapour Pressure** Surface area and volume of the liquid do **not** affect the vapour pressure: 	ext{Vapour pressure} 
e f(	ext{surface area}) 	ext{Vapour pressure} 
e f(	ext{volume}) **Conclusion** Vapour pressure depends only on: \boxed{	ext{Temperature} \quad 	ext{and} \quad 	ext{Nature of intermolecular forces}}

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A. Surface area only

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C. Volume of the liquid

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D. Atmospheric pressure

Question 3

At critical point, the compressibility factor Z equals:

Accepted Answer

Correct answer: C. 0.27 — At the critical point, the compressibility factor is defined as Z_c = \frac{P_c V_c}{R T_c}. For a van der Waals gas, substituting the critical constants P_c = \dfrac{a}{27b^2} , V_c = 3b , and T_c = \dfrac{8a}{27Rb} gives \begin{aligned} Z_c &= \frac{P_c V_c}{R T_c} \ &= \frac{\left(\dfrac{a}{27b^2}ight)(3b)}{R \left(\dfrac{8a}{27Rb}ight)} \ &= \frac{\dfrac{3a}{27b}}{\dfrac{8a}{27b}} \ &= \frac{3}{8} \ &= 0.375. \end{aligned} For most real gases, the experimentally observed value is approximately \boxed{Z_c = \frac{P_c V_c}{R T_c} \approx 0.27.}

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A. 1

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B. 0.5

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D. 2

Question 4

Which colligative property is independent of the nature of solute?

Accepted Answer

Correct answer: B. Boiling point elevation — 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 \Delta T_b = K_b \cdot m 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. \boxed{\Delta T_b = K_b \cdot m}

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A. Optical activity

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C. Refractive index

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D. Specific rotation

Question 5

For a spontaneous process at constant T and P, the condition is:

Accepted Answer

Correct answer: C. ΔG < 0 — At constant temperature and pressure, the criterion for spontaneity is \Delta G < 0, where the Gibbs free energy change is defined as \Delta G = \Delta H - T\,\Delta S. 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 \boxed{\Delta G < 0.}

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A. ΔH < 0

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B. ΔS > 0

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D. ΔU < 0

Question 6

The rate constant of a reaction at 25°C is k₁ and at 75°C is k₂. If Ea = 60 kJ/mol, what is k₂/k₁? (R = 8.314 J/mol·K)

Accepted Answer

Correct answer: C. 103 — **Step 1: Identify the Arrhenius two-temperature relation** The two-temperature form of the Arrhenius equation is \ln\!\left(\frac{k_2}{k_1}ight) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight) where T_1 = 25 + 273 = 298\ 	ext{K} and T_2 = 75 + 273 = 348\ 	ext{K} . **Step 2: Substitute the known values** \begin{aligned} \ln\!\left(\frac{k_2}{k_1}ight) &= \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight) \ 6pt] &= \frac{60000\ 	ext{J mol}^{-1}}{8.314\ 	ext{J mol}^{-1}	ext{K}^{-1}}\left(\frac{1}{298\ 	ext{K}} - \frac{1}{348\ 	ext{K}}ight) \ 6pt] &= 7215.3 	imes \left(3.356	imes10^{-3} - 2.874	imes10^{-3}ight)\ 	ext{K}^{-1} \ 6pt] &= 7215.3 	imes 4.820	imes10^{-4} \ 6pt] &\approx 4.63 \end{aligned} **Step 3: Solve for k_2/k_1 ** \frac{k_2}{k_1} = e^{4.63} \approx 103 **Result:** \boxed{\dfrac{k_2}{k_1} \approx 103}

Answer

A. 10.3

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B. 25.8

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D. 258

Question 7

Which of the following is an extensive property?

Accepted Answer

Correct answer: C. Entropy — **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.** \begin{aligned} 	ext{Entropy } (S) &: 	ext{extensive} \quad (	ext{depends on amount of substance}) \ 	ext{Refractive index } (n) &: 	ext{intensive} \ 	ext{Surface tension } (\gamma) &: 	ext{intensive} \ 	ext{Density } (ho) &: 	ext{intensive} \end{aligned} **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. \boxed{S = 	ext{Entropy is an extensive property.}}

Answer

A. Refractive index

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B. Surface tension

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D. Density

Question 8

For an ideal gas undergoing isothermal expansion, the work done by gas is:

Accepted Answer

Correct answer: B. nRT ln(Vf/Vi) — **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: W = nRT \ln\!\left(\frac{V_f}{V_i}\right) 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: \frac{V_f}{V_i} > 1 \implies \ln\!\left(\frac{V_f}{V_i}\right) > 0 Therefore, the work done by the gas is positive: \boxed{W = nRT \ln\!\left(\frac{V_f}{V_i}\right) > 0}

Answer

A. Zero

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C. nRT ln(Vi/Vf)

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D. ΔU

JEE Chemistry - MCQ Practice Questions