HBSE Class 12 Chemistry Question Paper 2022 Answer Key

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HBSE Class 12 Chemistry Question Paper 2022 Answer Key

SET-A,B,C,D (Subjective Questions)

Q1. If the density of lake water is 1.25 g mL-¹ and contains 92 g of Na+ ions per kg of water. Calculate the molality of Na+ ions in lake.

Ans. Mass of Na+ ions = 92 g

Molar mass of Na+ ions = 23 g/mol or 23 g mol-¹

Number of moles of Na+ ions = 92g/23gmol-¹ = 4 

Mass of water = 1 kg

Molality = (number of moles of sodium ions) ÷ 

Mass of water (in kg) 

Molality= 4/1 = 4 M 


Q2. Calculate the half life of a first order reaction from rate constant 200 sec-¹. 

Ans. t½ = 0.693/k = 0.693/200 = 0.003465 second or 3.4 × 10-³ second


Q3. What is Gabriel Phthalimide synthesis reaction ? 

Ans. The reaction of phthalimide with ethanolic KOH gives potassium salt of phthalimide. It is then heated with alkyl halide to form N-alkyl phthalimide. Alkaline hydrolysis (or treatment with hydrazine) gives a primary amine. Secondary or tertiary amines are not obtained as over alkylation is avoided. Aromatic primary amines cannot be prepared as aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide. 


Q4. Atoms of element B form hep lattice and those of the element A occupy 2/3 of tetrahedral voids. What is the compound formed by A and B ? 

Ans. The number of tetrahedral voids formed is equal to twice the number of atoms of element B. But only 2/3 of these tetrahedral voids are occupied by atoms of element A. Therefore the ratio of number of atoms of A and B is 2x(2/3) : 1 = 4 : 3

The formula of compound is A4B3


Q5. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol-¹. Find the molar mass of the solute. 

Ans. Here , given that kf = 5.12 K kg mol-¹ 

Mass of solvent, W1 = 50 g 

Mass of solute, W2 = 1 g 

∆Tf = 0.4 K

We need to find M2 , the molar mass of solute.

From the relation , we have

∆Tf = (kf×W2×1000) ÷ (W1×M2

M2 = (kf×W2×1000) ÷ (W1×∆Tf) 

      = (5.12 x1×1000) ÷ (50×0.4)

      =5120 ÷ 20 = 256 g

Molar mass of solute (M2) = 256 g


Q6. Explain giving reasons transition metals generally form coloured compounds. 

Ans. Colour of transistion metal is due to the excitation of an electron from a lower energy d-orbital to a higher energy d orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary color of the light absorbed (whose frequency lies on the visible region). 

The frequency of the light absorbed depends on the nature of the ligand. Transition metals form colored compounds due to the presence of vacant d-orbitals from the d-d transition of electrons which causes the color.


Q7. What happens when

(i) n-butyl chloride is treated with alcoholic KOH ?

Ans. When n- butyl chloride reacts with alcoholic KOH, the product formed is butene. This reaction is known as hydrohalogenation. 

CH3–CH2–CH2–CH2–Cl + KOH(alc) Dehydration → CH3–CH2–CH2=CH2 + H2


(ii) Bromobenzene is treated with Mg in the presence of dry ether?

Ans. The reaction of bromobenzene with Mg in the presence of dry ether, the product of this reaction is Phenylmagnesium bromide. 

Ph–Br + Mg + dry ether → Ph–MgBr 


(iii) Ethylbromide is treated with Na in the presence of dry ether?

Ans. Butane is formed when ethyl bromide is treated with sodium metal in dry ether. When sodium metals are reacted with alkyl bromides in the presence of dry ether, alkanes are produced. This reaction is called Wurtz reaction. 

2C2H5Br + 2Na + dry ether → C4H10 + 2NaBr 


Q8. Write the mechanism of acid dehydration of ethanol to yield ethene. 

Ans. Step I : Protonation of oxygen atom of −OH group.

Step II : Loss of a molecule of water to form carbonium ion.

Step III : Deprotonation to form carbon carbon double bond.


Q9. A solution of Ni(NO3)2 is electrolysed between Platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the Cathode ? (Ni^58.7) 

Ans. Current, I = 5 amperes 

Time = 20 × 60 = 1200 second

Charge = current × time = 5 × 1200 = 6000 C 

According to the reaction, 

Na²+(aq.) + 2e- → Ni(s) 

Nickel deposit by (2×96487)C = 58.7 g 

Nickel deposit by 6000 C = (58.7×6000) ÷ (2×96487) = 1.825 g 

Hence 1.825g of nickel will be deposit at the cathode. 


Q10. What are lyophilic and lyophobic sols ? Give one example of each type. 

Ans. Lyophilic Sols– If dispersed phase (solid) tends to attract (i.e., like or love) dispersion medium (liquid), the resultant sol is termed as lyophilic sol. Examples : Sol of gum, gelatine, starch etc. are lyophilic sols.


Lyophobic Sols– On the other hand if dispersed phase tends to repel (i.e., dislike or hatred) dispersion medium, the resultant sol is termed as lyophobic sol. Examples : Sols of metals, metal hydroxides, metal sulphides etc. are lyophobic sols.


Q11.(a) Draw structure of 3-Hydroxy butanal. 




(b) Describe Cannizzaro reaction.

Ans. Aldehydes which do not contain hydrogen when treated with a concentrated solution of an alkali undergo self oxidation-reduction. As a result, one molecule of aldehyde is reduced to corresponding alcohol while the other molecule is oxidized to the corresponding acid. This reaction is called Cannizzaro reaction.


(c) Describe Cross Aldol Condensation reaction.

Ans. When aldol condensation is carried out between two different aldehydes and or ketones, it is called cross aldol condensation. If both the reactants contain α-hydrogen atoms, it gives a mixture of four products. If both the reactants contain α-hydrogens, four compounds are obtained as products. For example, ethanal and propanal react to give four products. 



(a) CH3COCH3 on reacting with the following will give what product ?

(i) NH2OH + CH3COCH3 →



(ii) HCN + CH3COCH3 →



(iii) NH2NHCONH2 + CH3COCH3 →




(b) Write the products of the following reactions :






Ans.(ii) C6H5CHO + HCl 


Q12.(a) Give the reason for bleaching action of Cl2

Ans. When chlorine reacts with water, it produces nascent oxygen. This nascent oxygen then combines with the coloured substances present in the organic matter to oxide them into colourless substances.

Cl2 + H2O → 2HCl + [O] 


Coloured substances + [O] → Oxidized colourless substance 


(b) What happens when SO2 is passed through an aqueous solution of Fe(Ill) salt ?

Ans. SO2 acts as a reducing agent when passed through an aqueous solution containing Fe(III) salt. It reduces Fe(III) to Fe(II) i.e., ferric ions to ferrous ions. 


(c) Describe anomalous behaviour of Nitrogen. 

Ans. Due to the small size of Nitrogen, it exhibits high ionization enthalpy, high electronegativity. 


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