Page No 364: - Chapter 12 Organic Chemistry class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 12.30:

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Answer:

Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as

Thus, the mass of the U-tube containing KOH increases. This increase in the mass of U-tube gives the mass of CO₂ produced. From its mass, the percentage of carbon in the organic compound can be estimated.

Question 12.31:

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Answer:

Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

Question 12.32:

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Answer:

Percentage of carbon in organic compound = 69 %

That is, 100 g of organic compound contains 69 g of carbon.

∴0.2 g of organic compound will contain  

Molecular mass of carbon dioxide, CO₂ = 44 g

That is, 12 g of carbon is contained in 44 g of CO₂.

Therefore, 0.138 g of carbon will be contained in  = 0.506 g of CO₂

Thus, 0.506 g of CO₂ will be produced on complete combustion of 0.2 g of organic compound.

Percentage of hydrogen in organic compound is 4.8.

i.e., 100 g of organic compound contains 4.8 g of hydrogen.

Therefore, 0.2 g of organic compound will contain

It is known that molecular mass of water (H₂O) is 18 g.

Thus, 2 g of hydrogen is contained in 18 g of water.

∴0.0096 g of hydrogen will be contained in  of water

Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic compound.

Question 12.33:

A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer:

Given that, total mass of organic compound = 0.50 g

60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.

60 mL of 0.5 M NaOH solution   H₂SO₄ = 30 mL of 0.5 M H₂SO₄

∴Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mL

Again, 20 mL of 0.5 MH₂SO₄ = 40 mL of 0.5 MNH₃

Also, since 1000 mL of 1 MNH₃ contains 14 g of nitrogen,

∴ 40 mL of 0.5 M NH₃ will contain  = 0.28 g of N

Therefore, percentage of nitrogen in 0.50 g of organic compound  = 56 %

Question 12.34:

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Answer:

Given that,

Mass of organic compound is 0.3780 g.

Mass of AgCl formed = 0.5740 g

1 mol of AgCl contains 1 mol of Cl.

Thus, mass of chlorine in 0.5740 g of AgCl

∴ Percentage of chlorine 

Hence, the percentage of chlorine present in the given organic chloro compound is .

Question 12.35:

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Answer:

Total mass of organic compound = 0.468 g [Given]

Mass of barium sulphate formed = 0.668 g [Given]

1 mol of BaSO₄ = 233 g of BaSO₄ = 32 g of sulphur

Thus, 0.668 g of BaSO₄ contains  of sulphur = 0.0917 g of sulphur

Therefore, percentage of sulphur   = 19.59 %

Hence, the percentage of sulphur in the given compound is 19.59 %.

Question 12.36:

In the organic compound CH₂=CH–CH₂–CH₂–C≡CH, the pair of hydridised orbitals involved in the formation of: C₂ – C₃ bond is:

(a) sp – sp²(b) sp – sp³(c) sp² – sp³ (d) sp³– sp³

Answer:

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are sp, sp, sp³, sp³, sp², and sp² hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of C₂-C₃ bond is sp – sp³.

Question 12.37:

In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

(a) Na₄[Fe(CN)₆] (b) Fe₄[Fe(CN)₆]₃ (c) Fe₂[Fe(CN)₆] (d) Fe₃[Fe(CN)₆]₄

Answer:

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

Hence, the Prussian blue colour is due to the formation of Fe₄[Fe(CN)₆]₃.

Question 12.38:

Which of the following carbocation is most stable?

(a)  (b)   (c) 

(d) 

Answer:

 is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups. An increased + I effect by three methyl groups stabilizes the positive charge on the carbocation.

Question 12.39:

The best and latest technique for isolation, purification and separation of organic compounds is:

(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography

Answer:

Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.

Question 12.40:

The reaction:

is classified as :

(a) electrophilic substitution (b) nucleophilic substitution

(c) elimination (d) addition

Answer:

It is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (OH^–) with a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH₃CH₂I to form ethanol.