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CBSE 12th Chemistry 2017 Guess Paper

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CBSE 12th Chemistry 2017 Guess Paper By 4ono.com

TIME - 3HR. | QUESTIONS - 26

THE MARKS ARE MENTIONED ON EACH QUESTION _________________________________________________________________________

SECTION – A

Q.1. What is the basicity of 𝑯𝟑𝑷𝑶𝟒? 1 mark Ans. 𝐻3𝑃𝑂4 has three ionisable hydrogen atoms. Hence, its basicity is 3.

Q.2. Write the IUPAC name of the given compound: 1 mark

Ans. 2,4,6 – tribromaniline

Q.3. A delta is formed at the meeting point of sea water and river water. Why? 1 mark Ans. River water is colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water the electrolytes present in sea water coagulate the colloidal solution of clay resulting in its deposition with the formation of delta.

Q. 4. Is (−𝑪𝑯𝟐 − 𝑪𝑯−)𝒏 a homopolymer or a copolymer. 1 mark

Ans. It is homopolymer.

Q.5. What is Tollens’s reagent? Write one usefulness of this reagent. 1 mark

Ans. Tollens’s reagent is an ammonia Cal silver nitrate solution. Tollens’s reagent is used to test an aldehyde. Both aliphatic and aromatic aldehydes reduce Tollens’s reagent.

SECTION – B

Q.6. Describe the following: 2 marks (i) Tyndall effect (ii) Shape-selective catalysis

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Ans. (i) Tyndall effect: The phenomenon of scattering of light by colloidal particles as a result of which the path of beam becomes visible is called Tyndall effect.

(ii) Shape-selective catalysis: The catalytic reaction which depend upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis.

Q.7. What are biodegradable and non-biodegradable detergents? Give one example of each class. 2 marks

Ans. Biodegradable detergents: Detergents having straight hydrocarbon chains are easily decomposed by micro-organism are called biodegradable detergents. Example: Sodium lauryl sulphate. Non-biodegradable detergents: The detergents containing branched hydro carbon chains are not easily degraded (or decomposed) by the micro-organisms are called Non-biodegradable detergents Example: Sodium 4 — (1, 3, 5, 7- tetramethylocyl) benzene sulphonate.

Q.8. Draw the structure of the following: 2 marks

(i) 𝑋𝑒𝐹2

(ii) 𝐵𝑟𝐹3

Ans.(i) Structure of 𝑋𝑒𝐹6

(ii) Structure of 𝐵𝑟𝐹3




Q.9. Complete the following chemical reaction equations: 2 marks (i) Cr2O7

2− + I− + H+⟶ (ii) MnO−4 + NO−2H+⟶ Ans. (i) 𝐶𝑟2𝑂7

2− + 𝐻+ + 6𝐼− ⟶ 2𝐶𝑟3+ + 7𝐻2𝑂 2𝐼−⟶ 𝐼2 + 2𝑒

− × 3 𝐶𝑟2𝑂72− + 6𝐼− +

14𝐻+ ⟶ 𝐶𝑟7𝑂72− + 𝐼2 + 3𝐻2𝑂

(ii) 2𝑀𝑛𝑂4− + 5𝑁𝑂2

− + 6𝐻+⟶ 2𝑀𝑛2+ + 5𝑁𝑂2− + 3𝐻2𝑂

Q.10. Differentiate between molecular structures and behaviors of thermoplastic and thermosetting polymers. Give one example of each type. 2 marks

Ans. Thermoplastics: Thermoplastics are linear polymers which can be repeatedly softened on heating and hardened on cooling and hence can be used again and again without any charge in chemical composition and mechanical strength. Example: Polythene and polypropene. Thermosetting: Thermosetting polymers, are permanently setting polymers. On heating in a mould, they get hardened and set and cannot be softened again. This hardening in heating is due to cross-linking between different polymer chain to give a three-dimensional network solid. Example: Bakelite.

SECTION – C

Q.11. How would you convert the following: 3 marks (i) Phenol to benzoquinone (ii) Propanone to 2-methylpropan-2 ol (iii) Propene to propan-2-ol

Ans.

(ii).




Q.12. (a) How do you convert the following: 3 marks

(i). Phenol to anisole (ii). Propan-2-ol to 2-methylpropan-2-ol (iii). Aniline to phenol

(a)

Ans. (i) Phenol to Anisole

(ii) Propan-2-ol to 2-methylpropan-2-ol

(iii) Aniline to Phenol




OR

(a) Write the mechanism of the following reaction:

𝟐𝑪𝑯𝟑𝑪𝑯𝟐𝑶𝑯𝑯+

→ 𝑪𝑯𝟑𝑪𝑯𝟐 − 𝑶 − 𝑪𝑯𝟐𝑪𝑯𝟑

(b)Write the equation involved in the acetylation of Salicylic acid.

Step 1

Step 2

Step 3

(C) Acetylation of Salicylic acid:




Q.13. (a) Write the mechanism of the following reaction.

𝑪𝑯𝟑𝑪𝑯𝟐𝑶𝑯 𝑯𝑩𝒓 → 𝑪𝑯𝟑𝑪𝑯𝟐𝑩𝒓 + 𝑯𝟐𝑶

(b) Write the equation involved in Reimer-Tiemann reaction. 3 Marks

Ans. (a) The reaction proceeds through.

Nucleophilic substitution bimolecular 𝑆𝑁2 mechanism, as shown: Inversion of

configuration takes place during the reaction.

(b)

Q.14. Write the IUPAC names of the following coordination compounds: 3 marks (i) [𝐂𝐫(𝐍𝐇𝟑)𝟑𝐂𝐥𝟑]

(ii) 𝐊𝟑[𝐅𝐞(𝐂𝐍)𝟔]

(iii) [𝐂𝐨𝐁𝐫𝟐(𝐞𝐦)𝟐]+, (𝐞𝐧 = 𝐞𝐭𝐡𝐲𝐥𝐞𝐧𝐞𝐝𝐢𝐚𝐦𝐢𝐧𝐞)

Ans. (i) Triamminetrichlorochromium (III) (ii) Potassium hex cyanoferrate (III) (iii) Dibromidobis (ethane -1, 2-diammine) Cobalt (III) ion.




Q.15. Write chemical equations for the following conversion: 3 marks

(i) Nitrobenzene to benzoic acid.

(ii) Benzyl chloride to 20phenylethanamine.

(iii) Aniline to benzyl alcohol.

Ans. (i) Nitrobenzene to benzoic acid.

(ii) Benzyl chloride to 2-phenylethanamine.

(iii) Aniline to benzyl alcohol




Q.16. Describe the following giving one example for each: 3 marks (i) Detergents (ii) Food preservatives (iii) Antacids

Ans. (i). Detergents: Detergents may be defined as ammonium, sulphate or hydrogen sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms. Example: The more common detergents are the sodium salts of long chain sulphonic acids.

(ii). Food Preservatives: Food preservatives are the substances used to prevent spoilage of food due to microbial growth during storage. The most common preservatives used are table salt, sugar, vegetable oils and sodium benzoate (C6H5COONa)

(iii). Antacids: These are those substances which neutralize the excess acid and bring the pH to appropriate level in stomach 𝐸𝑥𝑎𝑚𝑝𝑙𝑒: Magnesium hydroxide and sodium bicarbonate.

Q. 17. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 of benzene has a boiling point of 80.31 °C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10 °C and 𝑲𝒃 for benzene = 2.53 °C kg m𝒐𝒍–𝟏)3 marks

Ans. Here 𝑤2 = 1.25 𝑔, 𝑤1 = 99.0 𝑔,𝑀2 = 78.08𝑔 𝑚𝑜𝑙−1

∆𝑇𝑏 = (80.31 − 80.10)℃ = 0.21℃ 𝑜𝑟 0.21𝐾

∴ ∆𝑇𝑏 = 𝐾𝑏𝑚 = 𝐾𝑏×𝑊2×1000

𝑀2×𝑊1

∴ 𝑀2 =𝐾𝑏×𝑊2×1000

∆𝑇𝑏×𝑊1=2.53×1.25×1000

0.21×99

𝑀2 =3162.5

20.79= 152.11 ≈ 152𝑔 𝑚𝑜𝑙−1

Q. 18. (i) Which one of the following is a Polysaccharide: starch, maltose, fructose, glucose

(ii) Write one difference between 𝜶-helix and 𝜷-pleated sheet structures of protein.

(iii) Write the name of the disease caused by the deficiency of vitamin 𝑩𝟏𝟐. 3 marks




Ans. (i) Starch is a polysaccharide.

(ii) 𝛼 -hellix: Polypeptide-chcin forms all possible H-bonds by twisting into a right handed helical structure with -NH group of each amino acid residue. Hydrogen bonded to the >c = 0 group of an adjacent turn of helix.

𝛽-structure: All peptide chain are stretched Out to nearly extension and then laid side by side which are held together by intermolecular Hydrogen bond.

(iii) Pernicious anemia and scurvy.

Q.19. Account for the following:

(i) Primary amines (𝑹 − 𝑵𝑯𝟐) have higher boiling point than tertiary amines (𝑹𝟑𝑵).

(ii) Aniline does not undergo Friedel – Crafts reaction.

(iii) (𝑪𝑯𝟑)𝟐𝑵𝑯 is more basic than (𝑪𝑯𝟑)𝟑𝑵 in an aqueous solution. 3 marks

OR

Give the structure of A, B and C in the Following reactions:

(i) 𝐶6𝐻5𝑁𝑂2𝑆𝑛+𝐻𝐶𝑙→

𝑨 𝑁𝑎𝑁𝑂2 + 𝐻𝐶𝑙

273𝑘 𝑩

𝐻2𝑂→ 𝑪

(ii) 𝐶𝐻3𝐶𝑁𝐻20/𝐻

+

→ 𝑨 𝑵𝑯𝟑

∆

𝑩𝑩𝒓𝟐+𝑲𝑶𝑯→ 𝑪.

Ans. (i) Due to maximum intermolecular hydrogen bonding in primary amines (due to presence of more number of H-atoms) primary amines have high BP in comparison to tertiary amines.

(ii) Anilines does not undergo Fredel-crafts reaction due to Acid-Base reaction between basic compound. Salt is formed with 𝐴𝑙 𝐶𝑙3. As a result, N, of aniline acquires (+) ve charge hence act as strong deactivating group.

Q.20. What mass of NaCl (𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 = 𝟓𝟖. 𝟓 𝐠 𝐦𝐨𝐥−𝟏) be dissolved in 65 g of water to lower the freezing point by 𝟕. 𝟓℃ ? The freezing point depression constant, 𝐊𝐟, for water is 𝟏. 𝟖𝟔 𝐊 𝐤𝐠 𝐦𝐨𝐥−𝟏. Assume van’t Hoff factor for NaCl is 1.87. 3 marks




Ans. Here M2 = 58.5 g mol−1 , ω1 = 65 g, ω2 =?

∆Tf = 7.5℃, i = 1.87, Kf = 1.86 K kg mol−1

Using the formula, ∆Tf = iKfm

∆Tf = iKf×w2×1000

M2×w1

7.5 = 1.87×1.86×w2×1000

58.5×65

w2 =7.5×58.5×65

1.87×1.86×1000=28518.75

3478.2= 8.199g

(iii) In (𝐶𝐻3)3𝑁 there is maximum steric hindrance and least solvation but in (𝐶𝐻3)2𝑁𝐻 the solvation is more and the steric hindrance is less than in (𝐶𝐻3)3𝐻; Although +I effect is less, since there are two methyl group; dimethyl amine is still a stronger base than trimethyl amine.

OR

Ans. (i) 𝐶6𝐻5𝑁𝑂2𝑆𝑛+𝐻𝐶𝑙→ (𝐴)

𝐶6𝐻5𝑁𝐻2𝑁𝑎𝑁𝑂2 + 𝐻𝐶𝑙

273𝑘 (𝐵)𝐶6𝐻5𝑁2𝐶𝑙

𝐻2𝑂→ (𝐶)𝐶6𝐻5𝑂𝐻

(ii) 𝐶𝐻3𝐶𝑁𝐻20

𝐻+

→

𝐶𝐻3 − 𝐶𝑂𝑂𝐻 𝑁𝐻3

∆ (𝐵)𝐶𝐻3 − 𝐶𝑂𝑁𝐻2

𝐵𝑟2+𝐾𝑂𝐻→ (𝐶)𝐶𝐻3 − 𝑁𝐻2

(A)

Q. 21. Account for the following observations: 3 marks (i) 𝐩𝐊𝐛 for aniline is more than that for methylamine. (ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide. (iii) Aniline does not undergo Friedel-Crafts reaction. Ans. (i) In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in CH3NH2 + I- effect of CH3 increases the electron density on the N—atom. Therefore, aniline is a weaker base than methylamine and hence its pKb value is higher than that of methylamine.




(ii) Methylamine being more basic than water, accepts a proton from water liberating OH−ions.

These OH− ions combine with Fe3+ ions present in H2O to form brown ppt. of hydrated ferric oxide.

FeCl3⟶ Fe3+ + 3Cl−2Fe3+ + 6OH−⟶ 2Fe(OH)3. Hydrated ferric oxide (Brown ppt.) (iii) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt

C6H5NH2 + AlCl3⟶ C6H5N+H2 + AlCl

−3

Lewis base Lewis acid As a result, N of aniline acquires +ve charge and hence it acts a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel-Crafts reaction.

Q.22. Name the following compounds according to IUPAC system. 3 marks

(i) 𝐂𝐇𝟑 − 𝐂𝐇 − 𝐂𝐇𝟐 − 𝐂𝐇 − 𝐂𝐇− 𝐂𝐇𝟑

| | 𝐂𝐇𝟑 𝐎𝐇

(ii)

(iii) 𝐂𝐇 − 𝐂 = 𝐂 − 𝐂𝐇𝟐 − 𝐎𝐇

| | 𝐂𝐇𝟑 𝐁𝐫

Ans. (i) 5-methylhexane, -3-ol. (ii) 3-Bromocyclohexanol. (iii) 2-Bromo-3-Methylbut-2-en-1-ol.




SECTION – D

Q.23. (a)What is meant by undictated, bidentate and ambidentate ligands? Give two examples for each.

(b)Calculate the overall complex dissociation equilibrium constant for the 𝑪𝒖(𝑵𝑯𝟑)𝟒

𝟐+ ion, given that 𝜷𝟒 for this complex is 𝟐. 𝟏×𝟏𝟎𝟏𝟑. 4 marks

Ans. (a). A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atom is called undictated ligands, e.g., Cl− and NH3. A molecule or an ion which has only one donor atom and hence form two coordinate bond with the central metal atom is called a bidentate ligand, e.g., NH2 − CH2 −

CH2 − NH2 and O̅OC − COO̅.

A molecule or an ion which contains two donor atoms but only of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g., CN− or NC̅ ∶ and ∶ NO2̅̅̅̅ or ∶ ONO

−.

(b). (i) Cu2+ + NH3 ⇌ Cu(NH3)2+

𝑘1 =[Cu(NH3)]

2+

[Cu2+][NH3]

(ii). 𝐶𝑢(𝑁𝐻3)2+ + 𝑁𝐻3 ⇌ 𝐶𝑢(𝑁𝐻3)2 2+

𝑘2 =[𝐶𝑢(𝑁𝐻3)2]

2+

[𝐶𝑢(𝑁𝐻3)]2+[𝑁𝐻3]

(𝐢𝐢𝐢) β4 =[Cu(NH3)4]

2+

[(NH3)]4[Cu2+ ]

[𝛽4 = 𝑘1×𝑘2×𝑘3×𝑘4]

SECTION – E

Q. 24. (a) write a suitable chemical equation to complete each of the following transformations: 5 marks (i) Butan-1-o1 to butanoic acid (ii) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid (b) An organic compound with molecular formula 𝑪𝟗𝑯𝟏𝟎𝑶 forms 2,4-DNP derivative, reduces Tonen's reagent and undergoes Cannizzaro's reaction. On vigorous oxidation it gives 1,2-benzenedicarboxylic acid, Identify the compound.




Ans. (a) (i) 𝐶𝐻3 − 𝐶𝐻2 − 𝐶𝐻2 − 𝐶𝐻2 − 𝑂𝐻 𝐾2𝐶𝑟2𝑂7𝐻+

→ [𝑂]

𝐶𝐻3𝐶𝐻2𝐶𝐻2𝐶𝑂𝑂𝐻

Butanoic acid

(ii)

(b)




OR

(a) Give chemical tests to distinguish between (i) Propanol and propanone (ii) Benzaldehyde and acetophenone (b) Arrange the following compounds in an increasing order of their property as indicated: (i) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN) (ii) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) (iii) 𝑪𝑯𝟑𝑪𝑯𝟐𝑪𝑯(𝑩𝒓)𝑪𝑶𝑶𝑯, 𝑪𝑯𝟑𝑪𝑯 (𝑩𝒓)𝑪𝑯𝟐𝑪𝑶𝑶𝑯, (𝑪𝑯𝟑)𝟐𝑪𝑯 COOH (acid strength) Ans. (a) (i). Propanol and propanone: Propanal being an aldehyde reduces Tollen's reagent to Silver mirror but propanone being a ketone does not. 𝐶𝐻3𝐶𝐻2𝐶𝐻𝑂 + 2[𝐴𝑔(𝑁𝐻3)2]

+ + 3𝑂𝐻−⟶ 𝐶𝐻3𝐶𝐻2𝐶𝑂𝑂− + 2𝐴𝑔 ↓ +4𝑁𝐻3 +

2𝐻2𝑂 Propanal Silver mirror

𝐶𝐻3𝐶𝑂𝐶𝐻3𝑇𝑜𝑙𝑙𝑒𝑛′𝑠

𝑟𝑒𝑎𝑔𝑒𝑛𝑡 𝑁𝑜 𝑠𝑖𝑙𝑣𝑒𝑟 𝑚𝑖𝑟𝑟𝑜𝑟

(ii) Benzaldehyde and acetophenone: Acetophenone gives iodoform test with NaOI whereas benzaldehyde does not. 𝐶𝐻3𝐶𝑂𝐶6𝐻5 + 3𝑁𝑎𝑂𝐼 ⟶ 𝐶6𝐻5𝐶𝑂𝑂𝑁𝑎 + 𝐶𝐻𝐼3 + 2𝑁𝑎𝑂𝐻 Acetophenone yellow ppt (b) (i) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)

(ii) 4-methoxybenzoic acid < benzoic acid < 3, 4-dinitrobenzoic acid. (iii) 𝐶𝐻3𝐶𝐻2𝐶𝐻𝐶𝑂𝑂𝐻 > 𝐶𝐻3𝐶𝐻 − 𝐶𝐻2 𝐶𝑂𝑂𝐻 > 𝐶𝐻3 − 𝐶𝐻 − 𝐶𝑂𝑂𝐻 | | | 𝐵𝑟 𝐵𝑟 𝐶𝐻3




Q. 25. Assign reasons for the following: 5 Marks

(i) Sulphur vapour is paramagnetic.

(ii) Ammonia (NH3) has greater affinity for protons than phosphine (𝑷𝑯𝟑).

(iii) The negative value of electron gain enthalpy of fluorine is less than that of

chlorine.

(iv) 𝑺𝑭𝟔 is much less reactive than 𝑺𝑭𝟒.

(v) Of the noble gases only xenon is known to form well-established chemical

compounds.

Ans. (i) In vapour state Sulphur partly exists as 𝑆2 molecule which has two unpaired electrons in the antibonding 𝜋∗ orbitals like 𝑂2 and hence it exhibits paramagnetism.

(ii) When 𝑁𝐻3 or 𝑃𝐻3 accepts a proton, an additional N-H or P-H bond is formed.

𝐻3𝑁:+ 𝐻+⟶ 𝑁𝐻4

+𝑎𝑛𝑑 𝐻3𝑃:+ 𝐻+⟶ 𝑃𝐻4

+

Due to the bigger size of P than N, P-H bond is weaker than N-H bond. As a result,

𝑃𝐻3 has less tendency than 𝑁𝐻3 to accept a proton. Therefore, 𝑁𝐻3 has higher proton affinity than 𝑃𝐻3.

(iii) The electron gain enthalpy of fluorine is less negative than that of chlorine due to the small size of fluorine atom due to that there is strong repulsion of electron.

(iv) In 𝑆𝐹6, six F atom protect the sulphur atom from attack by the reagents to such an

extent that even thermodynamically most favorable reactions like hydrolysis do not

occur. But in 𝑆𝐹4, S is not sterically protected since it is surrounded by only four F

atoms, hence 𝑆𝐹4 undergoes hydrolysis.

(v) Except radon which is radioactive Xe has least ionization energy among noble gases and hence it is readily forms chemical compounds particularly with oxygen and fluorine.




OR

(a) Describe the favorable conditions for the manufacture of (i) ammonia by Haber's

process, and (ii) sulphuric acid by contact process.

(b) Draw the structures of the following:

(i) 𝑷𝑪𝒍𝟓(𝐠)

(ii) 𝑺𝟖(𝐠)

(iii) 𝑪𝑰𝑭𝟑 (𝐠)

Ans. (a) (i) Ona large scale, ammonia is manufactured by Haber's process

𝑁2(g) + 3𝐻2(g) ⇌ 2𝑁𝐻3 (g)

The optimum conditions for the production of ammonia are a pressure about 200

atm, a temperature of about 700 K and the use of catalyst such as iron oxide with

small amount of 𝐾2𝑂 and 𝐴𝑙2𝑂3

(ii) Low temperature about 720K and high pressure about 2 bar and a catalyst 𝑉2𝑂5 are favorable conditions for the manufacture of 𝐻2𝑆𝑂4 by contact process.

(b) (i) Structure of 𝑃𝐶𝑙5 (g)

(ii) Structure of 𝑆8 (g)




(iii) Structure of 𝐶𝐼𝐹3 (g)

Q.26. (a) Differentiate between molarity and molality for a solution. How does a change in temperature influence their values? 5 marks (b) Calculate the freezing point of an aqueous solution containing 10.50 g of 𝐌𝐠𝐁𝐫𝟐 in 200 g of water. (Molar mass of 𝐌𝐠𝐁𝐫𝟐 = 𝟏𝟖𝟒 𝐠) (𝐊𝐟 for water = 1.86 K kg 𝐦𝐨𝐥−𝟏)

Ans. (a). Molarity: It is defined as the no, of moles of solute dissolved in 1kg of solvent. Molarity: It is defined as the no, of moles of solute dissolved in 1 L of solution.

(b) Molecular mass of MgBr2 = 184 g i = 3 as MgBr2⟶Mg2+ + 2 Br−

∆Tf = iKf×WB×1000

MB×WA=1.86×10.50×1000×3

184×200= 1.592 k

Freezing point of solution (Ts) = 273 − 1.592 = 271.408 K

OR

(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.

(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. (𝐊𝐛 𝐟𝐨𝐫 𝐰𝐚𝐭𝐞𝐫 = 𝟎. 𝟓𝟏𝟐 𝐊 𝐤𝐠 𝐦𝐨𝐥−𝟏, 𝐌𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐍𝐚𝐂𝐥 = 𝟓𝟖. 𝟒𝟒 𝐠)




Ans. (a) Osmosis: The spontaneous flow of solvent molecules from a less concentrated solution to a more concentrated solution though semi permeable membrane.

Osmotic pressure: The minimum excess pressure that has to be applied on the solution to prevent the entry of solvent into the solution through SPM.

It is a colligative property because it depends upon, no. of solute particles not on the nature of particle. E.g., o. | MKCl and Na Cl has same osmotic pressure under same temp and pressure condition.

(b) Elevation is boiling point [𝑖 = 2 as NaCl ⇌ Na+ + Cl]

∆𝑇𝑏 =𝐾𝑏×1000×𝑊𝐵𝑊𝐴×𝑀𝐵

=0.512×1000×15×2

250×58.44=512×2

974= 1.051 k

Boiling point of water = 373 𝑘

373 + 1.051 = 374.051 𝑘.



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