ISC Chemistry 2010 Class-12 Previous Year Question Paper Step by step Solutions of Part-1 and Part-2 (section -A, B, C) . By the practice of ISC Chemistry 2010 Class-12 Solved Previous Year Question Paper you can get the idea of solving.

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## ISC Chemistry 2010 Class-12 Previous Year Question Paper Solved

-: Select Options :-

Part – I

Part – II Section – A

Part – II Section – B

Part – II Section – C

Maximum Marks: 70
Time allowed: 3 hours

• Answer all questions in Part I and six questions from Part II, choosing two questions from Section A, two from Section B and two from Section C.
• All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
• The intended marks for questions or parts of questions are given in brackets [ ].
• Balanced equations must be given wherever possible and diagrams where they are helpful.
• When solving numerical problems, all essential working must be shown.
• In working out problems use the following data:
Gas constant R = 1.987 cal deg-1 mol-1 = 8.314 JK-1 mol-1 = 0.0821 dm3 atm K-1 mol-1. 1L atm = 1 dm3 atm = 101.3 J.

### Part – I (20 Marks)(Answer all questions)

ISC Chemistry 2010 Class-12 Previous Year Question Paper Solved

Question 1.
(a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets: [5]
(benzoic acid, negative, positive, vapour, vapour pressure, benzal chloride, more, less, electropositivity, electronegativity, reducing, oxidising, basic, acidic, PCl5, SOCl2, phenol, water, ice.)
(i) When water freezes to ………….. the free energy of the system is ……….
(ii) The …………. pressure of an aqueous solution of 0.1 M cane sugar is ………… than that of pure water.
(iii) When benzaldehyde reacts with ……….. it forms ………. and POCl3.
(iv) An aqueous solution of a mixture of ammonium chloride and ammonium hydroxide is a ………. buffer solution with pH ………… than seven.
(v) Halogens are strong ……….. agents because of their high

(b) Complete the following statements by selecting the correct alternative from the choices given: [5]
(i) The hybridization of the iron atom in [Fe(CN)6]3- complex is:
1. sp3
2. d2sp3
3. sp3d2
4. dsp2

(ii) The product formed when aniline is warmed with chloroform and caustic potash is:
1. Phenyl chloride
2. Methyl isocyanide
3. Phenyl isocyanide
4. Nitro phenol

(iii) For a dissociated solute in solution the value of van’t Hoff factor is:
1. Zero
2. One
3. Greater than one
4. Less than one

(iv) The unit of equivalent conductance is:
1. ohm-1 cm2 equiv-1
2. ohm-1 cm2 gm-1
3. ohm cm2 equiv-1
4. ohm-1 mole-1

(v) An example of an intensive property is:
1. Number of moles
2. Mass
3. Volume
4. Density

(c) Answer the following questions: [5]
(i) The reduction potential of a metal X is -0.76 volts while that of Y is -2.38 volts. Which of the two metals is a stronger reducing agent? Give a reason for your answer.
(ii) The osmotic pressure of a 0.25 M urea solution is 2.67 atm. What will be the osmotic pressure of a 0.25 M solution of potassium sulphate?
(iii) Name the type of isomerism exhibited by lactic acid, CH3CH(OH)COOH giving a reason for your answer.
(iv) Write the relationship between Gibb’s free energy, enthalpy, entropy and the temperature of a system. What is this equation known as?
(v) The elevation of boiling point produced by dilute equimolal solutions of three substances are in the order A > glucose > B. Suggest a reason for this observation.

(d) Match the following: [5]

 (a) Chiral carbon atom (i) Nernst equation (b) Hexadentate (ii) Phenol (c) Electrochemical cells (iii) Entropy (d) Reimer Tiemann reaction (iv) EDTA (e) Second Law of Thermodynamics. (v) Polarised light

(a)

(i) ice, negative
(ii) vapour, less
(iii) PCl5, benzal chloride
(iv) basic, more
(v) oxidising, electronegativity

(b)

(i) 2
(ii) 3
(iii) 3
(iv) 1
(v) 4

(c)

(i) X is a stronger reducing agent because the lesser the value of reduction potential, more is the tendency to lose electrons and stronger is the reducing agent.
(ii) Potassium sulphate is an electrolyte and ionises to give three ions
$\left(\mathrm{K}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{K}+\mathrm{SO}_{4}^{2-}\right)$, i = 3
For 0.25 M urea, π = 2.67 atm i.e., CRT = 2.67 atm
For 0.25 M K2SO4
π = i CRT = i × 2.67 atm = 3 × 2.67 atm = 8.01 atm
(iii) Enantiomerism (optical isomerism), it has a chiral centre and possesses chirality (dissymmetry).

(iv) ∆G = ∆H – T∆S
Where  ∆G = Gibb’s free energy change
∆H = Enthalpy change
∆S = Entropy change
T = Temperature in Kelvin
This equation is called Gibb’s Helmholtz equation.
(v) It is because substance A undergoes dissociation and substance B undergoes association while glucose is a non-electrolyte and neither undergoes association nor dissociation.

(d)

(i) (c)
(ii) (d)
(iii) (e)
(iv) (b)
(v) (a)

### Part – II (50 Marks)

A nswer six questions choosing two from Section A, two from Section B and two from Section C.

Section – A

ISC Chemistry 2010 Class-12 Previous Year Question Paper Solved

Question 2.
(a)

(i) What is the mass of a non-volatile solute (molar mass 60) that needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution? [3$\frac { 1 }{ 2 }$] (ii) Show that the time required for the completion of 75% of a reaction of the first order is twice the time required for the completion of 50% of the reaction. [2$\frac { 1 }{ 2 }$] (b) Give reasons for the following:
(i) The density of ice is less than that of water. [2] (ii) A solution of potassium carbonate turns red litmus paper blue while that of potassium sulphate has no effect on litmus. [2] Answer 2:
(a)

(i) Let mass of non-volatile solute (w2) = w2 g
Molar mass of non-volatile solute (M2) = 60
Mass of solvent (water) w1 = 100 g
The molar mass of water M1 = 18
Let V.P. of pure water (p°) = 100 mm
VP. of solution (ps) = 100 – 25 = 75 mm
Now,

(b)

(i) In case of ice, the hydrogen bonding gives rise to the cage-like structure of water molecules in which each water molecule is tetrahedrally surrounded by four water molecules. The molecules are not so closely packed as they are in the liquid state. When ice melts, cage-like structure collapses, water molecules come closer. For the same mass of water, volume decreases and density increases, Thus, ice has a lower density than that of water.

(ii) When potassium carbonate undergoes hydrolysis, it forms a strong base KOH and a weak acid H2CO3.

As the base produced (KOH) is strong, the resulting solution is alkaline. Therefore, the solution turns red litmus blue.
Potassium sulphate upon hydrolysis forms potassium hydroxide (strong base) and sulphuric acid (strong acid). As both acid and base produced are strong, the resulting solution is neutral and it has no effect on litmus paper.

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