HSC SAMPLE PAPER-2 Chemistry

26. Section D

27. Attempt any Three of the following questions:

27. (i) In solution is . Calculate the of solution.

(ii) The solubility product of is . Calculate its solubility in and (Molar mass of )

28. Draw a neat diagram for the Haworth formula of glucopyranose. Give the industrial applications of enzyme catalysis.
29. Distinguish between rhombic sulphur and monoclinic sulphur.
30. Write structural formulae for:

(i) Pentane-1,4-diol

(ii) Cyclohex-2-en-1-ol.

(iii) -Nitrophenol

(iv) Salicylic acid

31. Explain the trends in atomic radii of -block elements.

28. [A]] Answer Key

29. Section A

1. (i) (a) Crystalline solids

(ii) (a) Anode

(iii) (c) Amine

(iv) (a) 0.54

(v) (c) 2,3-Dibromobutane

(vi) (b)

(vii) (d)

(viii) (c)

(ix) (b) Chlorine

(x) (c) in water

2. (i) The standard cell potential for the reaction is .

(ii) 2-Bromo-3-methylpent-3-ene

(iii) Chain growth polymers are polyacrylonitrile, polyvinylchloride, polythene, etc. (iv) Propanamide produces ethanamine by Hofmann bromamide degradation reaction

(v) Uses of chlorine dioxide are as follows:

(a) Bleaching agent for paper pulp and textiles.

(b) In water treatment.

(vi) (a) The decrease in the force of attraction exerted by the nucleus on the valence electrons due to the presence of electrons in the inner shells is called the shielding effect.

(b) As a result of the shielding effect, the effective nuclear charge experienced by the valence electron is less than the actual nuclear charge.

(vii)

(viii)

30. Section B

3. (i) Thiosulphuric acid, .

(ii) Peroxy monosulphuric acid,

4. (i) To promote innovative chemical technologies that reduce or eliminate the use or generation of hazardous substances in the design, manufacture, and use of chemical products.

(ii) The green chemistry helps to reduce capital expenditure to prevent pollution.

(iii) Green chemistry incorporates pollution, prevention practices in the manufacture of chemicals and promotes pollution prevention and industrial ecology.

(iv) Green chemistry is a new way of looking at chemicals and their manufacturing process to minimise any negative environmental effects.

Calculate: Using formula (i).

Quantity of electricity passed

Using formula (ii),

Number of moles of electrons passed

Number of moles of electrons passed through molten is 0.03 mol .

6. (i)

(ii)

7. Nitrogen atom of amines contains a lone pair of electrons which can be donated. Thus, amines act as bases and nucleophiles.

e.g.,

(i) The reaction of ethylamine with dilute hydrochloric acid results in the formation of ethyl ammonium chloride.

8. Sulphur dioxide is angular with bond angle of

The S-O double bond arises from bonding. It is a resonance hybrid of two canonical forms.

9. The enthalpy change accompanying the dissociation of all molecules in one mole of gas phase substance into gaseous atoms is called enthalpy of atomisation.

For example: Atomisation of methane molecule.

10. As we move from to , the atomic and ionic radii of trivalent lanthanoid show steady decrease because with increase in atomic number, the nuclear charge increases and electrons are added to shell which show poor shilding effect. As a result there is gradual increase in nuclear charge and decrease in radii. This is known as lanthanoid contraction.
11. (i) The negatively charged acetate ion (i.e., the conjugate base of acetic acid) gets destabilised by + I effect of group.

(ii) Lesser is the stabilisation of the conjugate base, weaker is the acid.

(iii) In formate ion, there is no such destabilisation effect.

Thus, formic acid is stronger than acetic acid.

31. 2. (i) Wurtz-Fitting Reaction:

(ii) Fitting reaction:

13. The conjugate base of -nitrophenol is better resonance stabilised due to six resonance structures compared to the five resonance structures of the conjugate base of phenol. The resonance structure has a negative charge on only electgronegative oxygen atoms. Hence, -nitrophenol is a stronger acid than phenol.
14. (i) At the boiling point of a liquid, its vapour pressure is equal to .

(ii) In order to reach boiling point the solution and solvent must be heated to a temperature at which their respective vapour pressures attain .

(iii) At any given temperature the vapour pressure of a solution is lower than that of pure solvent. Hence, the vapour pressure of solution needs a higher temperature to reach than that needed for vapour pressure of solvent. Therefore, vapour pressure lowering causes a rise in the boiling point of a solution.

Section C

15. Ketones are prepared by reacting nitriles with Grignard reagent in dry ether as solvent followed by acid hydrolysis.

(Ethanenitrile)

16. (i) Using formula (i),

(ii) Using formula (ii),

Number of unit cells in volume of metal

Number of unit cells in of

(a) Radius of Al atom ( ) is 125 pm.

(b) Number of unit cells in of is .

17.

Factor
Kinetics	2nd order	1st order
Molecularity	Bimolecular	Unimolecular
Number of steps	One step	Two step
Bond makingand bondbreaking	Simultaneous	First the bond in thereactant breaks andthen a new bond inproduct is formed
Transition state	One step, onetransition state	Two steps, twotransition state
Directionof attack ofnucleophile	Only back sideattack	Back side attack andfront side attak
Nucleophile	StrongNucleophilefavourable.	Weak Nucleophilefavourable.

18. Molar mass of urea

Molar mass of water

Now, using formula,

19. Step 1:

33. Step 2:

20. (i) The oxidation state of vanadium is +3 .

(ii) Valence shell electronic configuration of free metal ion,

(iii) The number of ligands is 4. Therefore, the number of vacant metal ion orbitals required for bonding with ligands must be four.

(iv) Four orbitals on metal available for hybridisation are one and three . The complex is tetrahedral.

(v) The four metal ion orbitals for bonding with ligands are derived from the hybridization.

(vi) Four vacant hybrid orbitals of overlap with four orbitals of ions.

(vii) Configuration after complex formation would be

(viii) The complex has two unpaired electrons. The structure of

21. The integrated rate law for the first order reaction is given by the equation

Where initial concentration of the reactant at The concentration falls to , at time from the start of the reaction. The concentration of the reactant falls to at time .

Therefore,

So, the equation can be written as

34. Hess’s law of constant heat summation:

Hess’s law of constant heat summation states that, “Overall the enthalpy change for a reaction is equal to sum of enthalpy changes of individual steps in the reaction”

35. For example:

The enthalpy changes for a chemical reaction is the same regardless of the path by which the reaction occurs. Hess’s law is a direct consequence of the fact that enthalpy is state function. The enthalpy change of a reaction depends only on the initial and final states and not on the path by which the reaction occurs.

To determine the overall equation of reaction, reactants and products in the individual steps are added or subtracted like algebraic entities.

Consider the synthesis of .

(i)

(ii)

(iii)

The sum of the enthalpy changes for steps (1) and (2) is

Natural Rubber	Neoprene
Monomer;

equal to enthalpy change for the overall reaction.

36. Application of Hess’s Law:

The Hess’s law has been useful to calculate the enthalpy the enthalpy changes for the reactions with their enthalpies being not known experimentally.

23. A reaction which has higher order true rate law but experimentally found to behave as first order are called pseudo first order reaction. For example: Hydrolysis of sucrose.

is an example of pesudo first order reaction, because water takes part in the reaction the true rate law rate indicates that the reaction must be second order. Similarly to the hydrolysis of ester, . It is constant and the rate law became rate Thus, the second order true law is converted, to first order rate law.

25. (i) Every oxidation reaction needs to be accompanied by a reduction reaction.

(ii) The occurrence of only oxidation or only reduction is not possible.

(iii) In galvanic cell oxidation and reduction occur simultaneously.

(iv) The potential associated with the redox can be experimentally measured. For the measurement of potential two electrodes need to be combined together where the redox reaction occurs.

Hence, it is impossible to measure the potential of a single electrode.

26. (i)

37. Section D

27. (i)

(ii)

38. Calculation:

From formula (i).

From formula (ii).

(ii) The solubility product of is:

The solubility in molar solubility in molar mass

28.

( – D- (+) – Glucopyranose) ( -D- (+) – Glucopyranose) Some examples of industrial application of enzyme catalysis are:

(i) Conversion of glucose to sweet-tasting fructose, using glucose isomerase.

(ii) Manufacture of new antibiotics, using pencillin G acylase.

(iii) Manufacture of laundry detergents, using proteases.

(iv) Manufacture of esters used in cosmetics, using genetically engineered enzyme.

29.

30. (i)

Pentane-1, 4-diol

(ii)

Cyclohex-2-en-1-ol

(iii)

(iv)

31. (i) Atomic radii of the elements of the transition series decrease gradually from left to right.

(ii) As we move across a transition series from left to right, the nuclear charge increases by one unit at a time.

(iii) The last filled electron enters a penultimate subshell. However, orbitals in an atom are less penetrating or more diffused and, therefore electrons offer smaller screening effects.

(iv) The result is that effective nuclear charge also increases as the atomic number increases along with a transition series. Hence the atomic radii gradually decrease across a transition series from left to right.

Trends in atomic radii of -block elements