HBSE Class 12th Chemistry Solved Question Paper 2021
HBSE Class 12 Chemistry Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Chemistry 2021. HBSE 12th Question Paper Download 2021. HBSE Class 12 Chemistry Paper Solution 2021. Haryana Board Class 12th Chemistry Question Paper 2021 Pdf Download with Answer.
Q1. A compound is formed by two elements M and N. The element N forms CCP and atoms of M occupy 1/3 of tetrahedral voids. What is the formula of compound ?
Ans. Ratio of the number of atoms of M to that of N is M : N = (2/3) : 1 = 2 : 3.
Thus, the formula of the compound is M2N3.
Q2. Concentrated HNO3 used in laboratory work is 68% HNO3 by mass in aqueous solution. What should be the molarity of such a sample of acid if density of solution is 1.504 g mL-¹ ?
Ans. 68% HNO3 by mass means
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of HNO3 = 1+14+16×3 = 63 g mol-¹
Number of moles of HNO3 = W/M = 68g/63gmol-¹ = 1.079 mol
Density of solution = 1,504 g mL-¹
Volume of solution = Mass/Density = 100g/1.504g mL-¹ =66.5 mL or 0.0665 L
Molarity = Moles of solute/Volume of solution in litres = 1.079 mol ÷ 0.0665 L = 16.23 mol L-1 or 16.23 M
The molarity should be 16.23 mol L-¹ or 16.23 M.
Q3. Calculate the mass of a non-volatile (Molar-mass 40 g mol-¹) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Ans. Let p be the vapour pressure of pure octane. The vapour pressure of solution will be 80/100 p = 0.8 p
Molar mass of solute (M) and octane (m) are 40 g/mol and 114 g/mol respectively. Mass of octane, w is 114 g.
(p-p’)/p = Wm/Mw
(p-0.8p)/p = (W×114)/(40×114)
W = 8g
Hence, 8g of solute are required.
Q4. Calculate the half life of a first order reaction from rate constant 500 sec-¹.
Ans. t½ = 0.693/k = 0.693/500 = 1.3 × 10-³ second
Q5. What is Hoffmann bromamide degradation reaction ?
Ans. When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine. This reaction involving degradation of amide and is popularly known as Hoffmann bromamide degradation reaction.
R–CO–NH2 + Br2 + 4NaOH → R–NH2 + Na2CO3 + 2NaBr + 2H2O
Q6. A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is mass of copper deposited at Cathode ? (Cu^63.5)
Ans. Time (t) = 10 min = 10 × 60 = 600 sec
Current I = 1.5 A
Charge = time × current = t × I = 600 × 1.5 = 900 C
Now, Mass of copper deposited = (Molar mass×Charge) ÷ (electrons transferred×Faradays constant) = (63.5×900) ÷ (2×96500) = 0.296 g
Therefore, the mass of copper deposited at the cathode is 0.296 g.
Q7. What is difference between multimolecular and macromolecular colloids ? Give one example of each.
Ans. Multimolecular Colloids– When a large number of atoms or small molecules (having diameters of less than 1nm) of a substance combine together in a dispersion medium to form aggregates having size in the colloidal range, the colloidal solutions thus formed are called multimolecular colloids. The species (atoms or molecules) constituting the dispersed particles in multimolecular colloids are held together by Vander Waals’ forces. The gold sol, sulphur sol etc. are some examples of multimolecular colloids.
Macromolecular Colloids– Certain substances form large molecules whose dimensions are comparable to those of colloidal particles. Such molecules have very high molecular masses and are termed as macromolecules. When such substances are dispersed in suitable dispersion medium, the resulting colloidal solutions are known as macromolecular colloids. Thus, in macromolecular colloids, the dispersed particles are themselves large molecules having very high molecular masses. For example colloidal dispersion of naturally occurring macromolecules such as starch, proteins, gelatin, cellulose, nucleic acids etc. are macromolecular colloids.
Q8. Explain giving reasons transition metals and their many compounds act as good catalyst.
Ans. Transition metals and their many compounds act as good catalyst. Transition metals show variable oxidation states and forms complexes. They form unstable intermediate compounds. They provide a new path with lower activation energy of reaction. They also provide a suitable surface for the reaction to occur. Transition metals have d and s orbitals to form these bonds.
Q9. What happens, when :
(A) Ethyl Chloride is treated with aqueous KOH.
Ans. C2H5Cl + aqueous KOH → C2H5OH + KCl
Ethanol (C2H5OH) is formed.
(B) Methyl bromide is treated with Na in presence of dry ether.
Ans. When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.
CH3–Br + 2Na + Br–CH3 + Dry Ether → CH3–CH3 + 2NaBr
(C) Methyl Chloride is treated with KCN.
Ans. When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.
CH3–Cl + KCN → CH3–CN + KCl
Q10. Write the Mechanism of hydration of ethene to yield ethanol.
Ans. Three steps are involved in the mechanism of hydration of ethene to yield ethanol.
Step-I The electrophillic attack of hydronium ion protonates ethene to form carbocation.
Step-II Water (as a nucleophile) attacks carbocation.
Step-III Deprotonation gives ethanol.
Q11.(i) H2S is less acidic than H2Te. Why ?
Ans. H2S is less acidic than H2Te because as we move down the group,the bond dissociation enthalpy decreases i.e it becomes easy to dissociate or break the bonds(due to the increasing size down the group).So,with this concept in terms of hydrides,it becomes easy for H2Te to remove H easy as H+ because of its low bond dissociation enthalpy(as mentioned before). Therefore, H2Te is more acidic.
(ii) Mention the conditions required to Maximise the yield of ammonia.
Ans. Ammonia is prepared using the Haber’s process. High pressure on the reaction at equilibrium favors the shift of the equilibrium to the right. A pressure of about 200atm will be suitable for the higher yield of ammonia. Temperature ~ 700k.
(iii) Describe anomalous behaviour of Oxygen.
Ans. The reasons are Small size, Higher electronegativity, Non-availabiity of d-orbitals.
Oxygen is a gas while others are solids at room temperature. Oxygen is a non-metal. Sulphur is non-metallic while others exhibit metallic character. Due to small size and high E.N, oxygen form pπ−pπ multiple bonds with elements having similar size. Oxygen is paramagnetic while others are diamagnetic.
(i) Which form of sulphur shows paramagnetic behaviour ?
(ii) Complete the following equations :
(a) Cu + 4HNO3 (conc.)→
Ans. Cu(NO3)2 + 2NO2 + 2H2
(b) 2NaOH + SO2 →
Ans. Na2SO3 + H2O
(c) NH3 + 3Cl2 (excess) →
Ans. NCl3 + 3HCl
Q12.(i) Draw structure of 4-Oxopentanal.
(ii) Describe Hell-Volhard-Zelinsky reaction.
Ans. This reaction involves alpha bromination of carboxylic acids. Carboxylic acids react with chlorine or bromine in presence of small amount of red phosphorous to give alpha halo carboxylic acids.
CH3–CH2−COOH + Br2/P → CH3–CHBr–COOH + Br2/P → CH3–CBr2–COOH + Br2/P → No further reaction
(iii) Describe Aldol condensation reaction.
Ans. Aldehydes having α hydrogen undergo self condensation on warming with dilute or mild base to give p-hydroxy aldahydes called aldols. This reaction is called aldol condensation. eg.
(i) Convert Benzene to acetophenone.
(ii) Convert Propanone to Propene.
(iii) Convert Ethanol to 3-Hydroxybutanal.
Q1. Percentage of empty space in a BCC arrangement is :
Ans. (C) 32%
Q2. The molality of pure water is :
Ans. (D) 55.5
Q3. For the given cell reaction : Cu/Cu²+ // Ag+/Ag
(A) Cu as cathode
(B) Ag as cathode
(C) Ag as oxidising agent
(D) None of the above
Ans. (B) Ag as cathode
Q4. Identify the order of reaction from given rate constant K=2.6×10-⁴ mol L-¹S-¹
(D) None of these
Ans. (B) Zero
Q5. According to Hardy-Schulze rule, which of the following has highest flocculating power ?
Ans. (A) Al³+
Q6. In the following compounds which has minimum boiling point ?
Ans. (D) H2S
Q7. In the following strongest reducing agent is :
Ans. (B) BiH3
Q8. Which of the following ion is colourless in aqueous solution ?
Ans. (D) Sc³+
Q9. Which Metal has highest density ?
Ans. (B) Os (Osmium)
Q10. What is the Co-ordination number in the [Co(NH3)6]Cl3 Compound ?
Ans. (C) 6
Q11. The Oxidation number of chromium in [Cr(H2O)6]Cl3 is :
Ans. (B) +3
Q12. Organic compound which shows complete stereochemical inversion during SN² reaction :
(D) None of the above
Ans. (A) CH3–Cl
Q13. Which of the following is most acidic ?
Ans. (D) M-Chlorophenol
Q14. Williamson Synthesis is used to prepare :
Ans. (D) Ether
Q15. IUPAC name of Acetic acid :
(A) Ethanoic acid
(D) Methanoic acid
Ans. (A) Ethanoic acid
Q16. Which of the following is a 3⁰ -amine ?
(B) Triethyl amine
Ans. (B) Triethyl amine
Q17. Which base is present in DNA but not in RNA ?
Ans. (A) Thyamine
Q18. In the following which is not a Monosaccharide sugar ?
Ans. (D) Maltose
Q19. What is Schottky defect ?
Ans. It is a point defect in which an atom or ion is missing from its normal site in the lattice.
Q20. What role does the Molecular interaction play in a solution of alcohol and water ?
Ans. To Break the Hydrogen Bond
Q21. What is a weak electrolyte ?
Ans. A weak electrolyte is a solution in which only a small fraction of the dissolved solute exists as ions.
Q22. Define collision frequency.
Ans. Collision frequency describes the rate of collisions between two atomic or molecular species in a given volume, per unit time.
Q23. What is observed when electric current passed through a colloidal sol.
Ans. When electric current is passed through a colloidal sol, the colloidal particles move towards oppositely electrode, where they lose their charge and are coagulated. This process is known as electrophoresis.
Q24. What are Ligands ?
Ans. Ligands is an ion or molecule that binds to a central metal atom to form a coordination complex.
Q25. What is Diazotisation reaction ?
Ans. The conversion of primary amine aromatic amines into diazonium salts is known as diazotisation.
Q26. What are the expected products of hydrolysis of Sucrose ?
Ans. The two products of hydrolysis of Sucrose are glucose and fructose.
Q27. Solids have ………… volume and …………. shape.
Ans. Definite, Definite
Q28. The sum of the powers of the concentration of the reactants in rate law called ………….
Ans. Order of reaction
Q29. The general electronic configuration of inner transition elements is ………….
Ans. (n-2)f¹-¹⁴ (n-1)⁰-¹ ns²
Q30. ………….. ions produced from complex compound K4[Fe(CN)6] ?
Q31. CH3Br + AgF → ………. + AgBr
Q32. CH3CH2OH + H2SO4 + 413K →
Q33. ………….. is the IUPAC name of the compound CH3CH(CH3)CH2CH2CHO.
Q34. …………….. is the IUPAC name of the complex compound K3[Cr(C2O4)3].
Ans. Potassium Trioxalatocromate(III)
Q35. …………. is the deficiency disease of Vitamin ‘D’.