Haryana Board (HBSE) Class 12 Chemistry SAT-1 Question Paper 2025 Answer Key. Get the Students Assessment Test (SAT) Class 12 Chemistry question paper with complete solution, accurate answer key, and expert preparation tips. The Haryana Board of School Education (HBSE) conducts SAT as an important assessment for Class 12 students. Best resource for Haryana Board Class 12 SAT Chemistry exam practice, quick revision, and scoring better marks.
HBSE Class 12 Chemistry SAT-1 Question Paper 2025 Answer Key
Instructions :
• All questions are compulsory.
• Questions (1-9) carry 1 mark each.
• Questions (10-12) carry 2 marks each.
• Questions (13-14) carry 3 marks each.
• Question (15) case study, carry 4 marks.
• Questions (16-17) carry 5 marks each.
1. 6.022 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of solution is :
(A) 0.1 M
(B) 0.02 M
(C) 0.01 M
(D) 0.001 M
Answer – (C) 0.01 M
Number of moles = Number of molecules / Avogadro’s number = 6.022 × 1020 / 6.022 × 1023 = 10–3 mol
Molarity = Number of moles / Volume in liters = 10–3 / 0.1 L = 0.01 M
2. If molality of a dilute solution is doubled, the value of molal depression constant will be :
(A) Halved
(B) Tripled
(C) Doubled
(D) Unchanged
Answer – (D) Unchanged
3. The direction of current in the Daniel cell, when zinc and copper electrodes are connected is :
(A) From copper to zinc in the cell
(B) From copper to zinc outside the cell
(C) From zinc to copper outside the cell
(D) From zinc to copper in the cell
Answer – (B) From copper to zinc outside the cell
4. The charge required to deposit 2 moles of Ag when current is passed through molten AgNO3 :
(A) 1 F
(B) 2 F
(C) 3 F
(D) 5 F
Answer – (B) 2 F
Q = n × F = 2 moles × 1 F = 2 F
5. If the rate constant of a reaction is 0.9 mol / L / s, then it will be a …………. order reaction.
Answer – Zero order
6. A (g) + B (g) → Product is an elementary reaction, if pressure of both A and B is increased three times of initial pressure, the rate of forward reaction will be …………….. of previous rate.
Answer : For gases, concentration is proportional to pressure (from the ideal gas law,
P = nRT/V, so [A] = PA/RT
If the pressure of both A and B is increased by a factor of 3, their concentrations also increase by a factor of 3
[A]new = 3[A]initial
[B]new = 3[B]initial
Rateinitial = k[A][B]
Ratenew = k[A]new[B]new = k(3[A])(3[B]) = 9k[A][B]
so, the rate of forward reaction will be 9 times.
7. The van’t hoff factor (i) for a dilute solution of K3[Fe(CN)6] is …………….
Answer : K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3–
Total number of particles = 3 + 1 = 4
so, Van’t hoff factor (i) = 4
8. For an elementary reaction X + Y → Z + W, the molecularity is ………….
Answer : Here two reactant molecules (X and Y) are involved simultaneously in the elementary step.
Molecularity = 2 (Bimolecular reaction)
9. Assertion (A) : Zn, Cd and Hg are considered as d – Block elements but not Transition Metals.
Reason (R) : Zn, Cd and Hg have partially filled d – orbitals.
(A) Both assertion and reason are correct and reason is the correct explanation of assertion.
(B) Both assertion and reason are correct but reason is not the correct explanation of assertion.
(C) Assertion is correct and the reason is wrong.
(D) Assertion is wrong and the reason is correct.
Answer – (C) Assertion is correct and the reason is wrong.
10. Calculate the mass of a non – volatile solute (Molar mass = 40 g/mol) that should be dissolved in 140 g of octane to lower its vapour pressure to 80%.
Answer : For a non-volatile solute, Raoult’s Law states :
(Po – P) / Po = xs
xs = 1 – P/Po = 1 – 0.8 = 0.2
Thus, the mole fraction of the solute xs = 0.2
The mole fraction of the solute is:
xs = ns / (ns + nw) = 0.2
ns = moles of solute
nw = moles of solvent
Mass of octane = 140 g, molar mass of octane = 114 g/mol
nw = mass of octane / molar mass of octane = 140 / 114 = 1.228 mol
ns / (ns + 1.228) = 0.2
ns = 0.307 mol
Molar mass of solute = 40 g/mol
Mass of solute = ns × molar pass = 0.307 × 40 = 12.28 g
11. Write down the differences between rate of reaction and specific reaction rate.
Answer : Rate of Reaction – The rate of reaction is the change in concentration of a reactant or product per unit time. It indicates how quickly a reaction is proceeding.
• Specific reaction rate – The specific reaction rate (k) is a proportionality constant in the rate law that relates the rate of a reaction to the concentrations of the reactants.
12. Explain why actinoid contraction is greater than that of lanthanoid contraction.
Answer – Actinoid contraction is greater than lanthanoid contraction because in the actinoid series the 5f orbitals are being filled, and these orbitals have a very poor shielding effect compared to the 4f orbitals in the lanthanoid series. As a result, the effective nuclear charge felt by the outer electrons increases more rapidly across the actinoid series, pulling the electron cloud closer to the nucleus and causing a stronger contraction in atomic and ionic sizes than what is observed in lanthanoids.
13. Resistance of a conductivity cell filled with 0.1 M KCI solution is 100 ohm. The resistance of the same cell when filled with 0.02 M KCI solution is 520 ohm. Calculate the conductivity and molar conductivity of 0.02 M KCI solution. The conductivity of 0.1 M KCI solution is 1.29 S/m.
Answer – Given : R0.1,M = 100 Ω, Κ0.1,Μ = 1.29 S m–1
Cell constant = кR = 1.29 × 100 = 129 m–1
For 0.02M: R = 520 Ω
k = 129/520 = 0.248 S m–1
Molar conductivity (SI) = k/c = 0.248/20 = 0.0124 S m2 mol–1 = 124 S cm² mol–1
14. The following results have been obtained during the kinetic studies of reaction :
A + B → C + D
| Experiment | [A] mol L–1 | [B] mol L–1 | Initial Rate formation of D (mol L–1 min–1) |
| I | 0.1 | 0.1 | 6.0 × 10–3 |
| II | 0.3 | 0.2 | 7.2 × 10–2 |
| III | 0.3 | 0.4 | 2.88 × 10–1 |
| IV | 0.4 | 0.1 | 2.40 × 10–2 |
Determine the rate law and the rate constant for the reaction.
Answer – Rate law: Rate = k[A][B]2
Order of reaction = 1 + 2 = 3
k = 6.0 L2 mol–2 min–1
15. CASE STUDY : An ideal solution may be defined as the solution which obeys Raoult’s law exactly over the entire range of concentration. The solutions for which vapour pressure is either higher or lower than predicted by Raoult’s law are called non ideal solutions. Non ideal solutions can show either positive or negative deviations from Raoult’s law depending on whether the A-B interactions in solution are stronger or weaker than A-A and B-B interactions.
Based on the above paragraph answer the following questions :
(I) What type of solution is formed when chloroform is mixed with acetone and why?
Answer – Negative deviation (chloroform + acetone form H-bonded A-B interactions stronger than A-A or B-B).
(II) What are the signs of ∆H and ∆V for a solution with positive deviation from ideal behaviour?
Answer : ΔΗmix > 0, ∆Vmix > for positive deviation.
(III) Name two factors on which the vapour pressure of liquid depends?
Answer – Vapour pressure depends on temperature and nature / intermolecular forces of the liquid.
16. Why transition elements :
(a) Show variable oxidation States?
Answer – Variable oxidation states because (n-1)d and ns orbitals have comparable energies, allowing variable numbers of electrons to participate in bonding.
(b) Acts as catalyst?
Answer – Act as catalysts due to variable oxidation states and ability to form intermediate complexes/adsorb reactants on their surfaces, lowering activation energy.
(c) Form coloured compounds?
Answer – Coloured compounds arise from d! → d transitions (and charge-transfer bands) in partially filled d-subshells.
(d) Have high enthalpy of atomization
Answer – High enthalpy of atomization due to strong metallic bonding from many unpaired d-electrons and extensive d-orbital overlap.
OR
Describe the preparation of potassium permanganate. How can you oxidize the following using acidified permanganate solution :
(i) Chloride ions into Chlorine
(ii) Sulphur Dioxide into Sulphate ions
(iii) Sulphide ion into Sulphur.
Answer – Preparation of Potassium Permanganate
MnO2 + KOH + O2 (fusion) → K2MnO4 (green)
K2MnO4 → KMnO4 (purple) by oxidation/disproportionation.
Oxidations with Acidified KMnO4 :
(i) CI– → CI2
2MnO4– + 10Cl– + 16H+ → 2Mn2+ + 5Cl2 + 8H2O
(ii) SO2 → SO42–
2MnO4– + 5SO2 + 2H2O → 2Mn2+ + 5SO42– + 4H+
(iii) S2– → S
2MnO4– + 5S2– 16H+ → 2Mn2+ + 5S + 8H2O
17. (i) Explain the electrolysis of an aqueous solution of AgNO3 using silver electrodes.
Answer – Electrolysis of Aqueous AgNO₃ with Silver Electrodes :
Electrolyte: AgNO3(aq) → Ag+ + NO3–
Electrodes: Both are silver (Ag)
At Cathode (−) : Ag+ ions gain electrons and deposit as metallic silver.
Ag+ + e– → Ag (s)
At Anode (+) : Silver electrode itself dissolves, releasing Ag+ ions.
Ag (s) → Ag+ + e–
(ii) Explain dry cell or Leclanche cell in detail.
Answer – A dry cell, also known as a Leclanche cell, is a type of primary electrochemical cell that converts chemical energy into electrical energy. It’s commonly used in devices like remote controls, toys, and flashlights due to its portability and convenience. The Leclanche cell was invented by Georges Leclanche in 1866, and the dry cell is a modified version of his invention.
Construction :
• Anode : A zinc container that serves as the negative electrode (anode).
• Cathode : A central carbon rod surrounded by a manganese dioxide (MnO2) and carbon mixture.
• Electrolyte : A moist paste of ammonium chloride (NH4Cl) and zinc chloride (ZnCl2).
OR
(i) Why is salt bridge used in electrochemical cell?
Answer – A salt bridge completes the circuit and maintains electrical neutrality by ion migration, preventing charge build-up and minimizing liquid junction potential; it does not allow mixing or electron flow.
(ii) Calculate the EMF of Cu – Ag cell when Cu rod is dipped in 0.1 M CuSO4 solution and Ag rod is dipped in 0.01 M AgNO3 solution at 298 K. (E°Cu2+/Cu = 0.34 V, E°Ag+/Ag = 0.80 V)
Answer – For the given reaction
Cu / Cu2+ (0.025M) // Ag+ (0.005M) / Ag
Ag+ / Ag act as cathode.
Cu / Cu2+ act as anode.
E°cell = E°cathode – E°anode = 0.80 – 0.34 = 0.46 V
The given reaction is
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag
From here n = 2
We can calculate the emf of cell by Nernst equation
Ecell = E°cell – 0.0591/n log[Cu2+] / [Ag+]2
= 0.46 – 0.0591/2 log [0.025] / [0.005]2
= 0.46 – 0.02995 log(103)
= 0.46 – 0.02995 × 3
= 0.46 – 0.08985
= 0.37 V