Sulphuric Acid Dalal Simplified ICSE Chemistry Class-10

Sulphuric Acid Dalal Simplified ICSE Chemistry Class-10 Solutions. Dalal Simplified ICSE Chemistry Sulphuric Acid by Dr Viraf and J Dalal for Class 10. Step by step Solutions of Sulphuric Acid by Dr Viraf and J Dalal of Concept of Simplified Dalal ICSE Chemistry.

Sulphuric Acid Dalal Simplified ICSE Chemistry Class-10

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ADDITIONAL QUESTIONS Sulphuric Acid Dalal Simplified ICSE 

Question 1.

State why sulphuric acid was called – ‘oil of vitriol’.
Answer:
Sulphuric acid was initially called ‘oil of vitriol ’.It was initially prepared by – distilling green vitriol [FeSO₄.7H₂O] and hence the name – ‘oil of vitriol’.

Question 2.

State how you would convert
(1) sulphur
(2) chlorine
(3) sulphur dioxide – to sulphur acid.
Answer:

1. S + 6HNO. [cone.] → 6NO₂ + 2H₂O + H₂SO₄
   
2. Cl₂ + SO₂ + 2H₂O → 2HCl + H₂SO₄
3. 3SO₂ + 2HNO₃ + 2H₂O → 3H₂SO₄ + 2NO

Question 3.

State the purpose of the ‘Contact process’.
Answer:
When sulphur is burnt in air, it bums with a pale blue flame forming sulphur dioxide and traces of sulphur trioxide.
S + O₂→ SO₂ [2S + 3O₂ → 2SO₄ (traces)]
Burning of sulphur or iron pyrites in oxygen is preferred to purified air since heat energy is wasted in heating the unreactive nitrogen component of the air.

Question 4.

In the Contact process

1. State how you would convert (a) sulphur (b) iron pyrites to sulphur dioxide in the first step of the Contact process.
2. State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.
3. State why the above catalytic oxidation {reaction supplies energy.
4. Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.
5. Give a reason why the catalyst mass is heated electrically – only initially.
6. State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.

Answer:

1 S + O₂ → 5- SO₂
4FeS₂ + 11O₂  → 5- 2Fe₂O₃ + 8SO₂

2 Catalytic oxidation of sulphur dioxide to sulphur trioxide.

[Above equation for the catalysed reaction is exothermic- hence supplies energy.]
Catalyst: Vanadium pentoxide [V₂O₅] or platinum [Pt],
Temperature: 450-500°C Pressure : 1 to 2 atmospheres
Conversion ratio: 98% of sulphur dioxide converted to sulphur trioxide.

3 The catalytic oxidation of SO₂ to SO₃ is an – exothermic reaction. Thus this reaction supplies energy in the form of heat.

4 Vanadium pentoxide is preferred – to platinized asbestos as a catalyst since it is comparatively – cheaper and less easily poisoned or susceptible to impurities.

5 The catalyst – mass is – only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500°C.

6 Even though sulphur trioxide is an acid anhydride of sulphuric acid – it is not directly absorbed in water to give sulphuric acid.
The reaction is highly exothermic resulting in production of – a dense fog of sulphuric acid particles which do not condense easily.
Hence sulphur trioxide vapours are – dissolved in cone, sulphuric acid to give oleum which on dilution with – the requisite amount of soft water in the dilution tank gives – sulphuric acid of the desired concentration [about 98%].

Question 5.

Give a reason why concentrated sulphuric acid is kept in air tight bottles.
Answer:
Concentrated sulphuric acid has a great affinity for water and as such it is a hygroscopic liquid. Being Hygroscopic, it absorbs moisture from the atmosphere and hence cone, sulphuric acid is kept in air tight bottles.

Question 6.

State the basic steps with reasons, involved in diluting a beaker of cone. H₂SO₄.
Answer:
For dilution of cone. H₂SO₄, the cone, acid is always added to water and water is never added to the cone, acid even though heat is evolved in both cases.
Reason: If water is added to cone. H₂SO₄ the heat produced is sufficient to spontaneously vaporise a part of the few drops of water added. This is because the amount of water is very small and bioling point of water is much lower than cone. H₂SO₄, which is in bulk. Due to this sudden vaporisation of water cone, acid tend to spurt out and cause serious injuries.
On the other hand, if cone. H₂SO₄ is added to water, the heat produced in this case can only raise the temperature of water slightly because water is an bulk. Thus, in this case spurting of the cone, acid is avoided.

Question 7.

Give reasons why dilute sulphuric acid:
(1) behaves as an acid when dilute.
(2) is dibasic in nature.
Answer:

1.  Acidic properties of sulphuric acid are due to the presence of – hydronium ions [H₃O⁺ ] formed when H₂SO₄ dissociates in aq. solution

2. Sulphuric acid dissociates in – aq. solution giving 2H⁺ ions per molecule – of  the acids .Hence its – basically is two.

[Basicity is the number of H⁺ ions formed by dissociation of one molecule of the acid in its aq. soln.]

Question 8.

Convert dil. H.SO, to –

1. Hydrogen
2. Carbon dioxide
3. Sulphur dioxide
4. Hydrogen sulphide
5. An acid salt
6.  A normal salt.

Answer:

1. Dil. H₂S0₄to sulphur dioxide by the action of any sulphide on dil. H₂SO₄

 

_{3. Dil. H2S04to sulphur dioxide by the action of any sulphide on dil. H2SO4}

Na₂SO₃ + H₂SO₄ → Na₂SO₄ + H₂O + SO₂

4. H₂SO₄ to hydrogen sulphide by the action of any sulphide on dil. H₂SO₄
FeS + H₂SO₄ (dil.) → FeSO₄ + H₂S↑

5. H₂S0₄to an acid by the action of insufficient strong base with excess of dil. H₂SO₄

Question 9.

Give equations for formation of two different acids from cone. H₂SO₄. State the property of sulphuric acid involved in the above formation.
Answer:

Property of Sulphuric acid involved in the formation of these acids: Cone. H₂SO₄ is a non-volatile acid.

Question 10.

Give equations for oxidation of cone. H₂SO₄ giving the oxidised products –

1. Carbon dioxide
2. Sulphur dioxide
3. Phosphoric acid
4. Copper (II) sulphate
5. Iodine
6. Sulphur respectively.

Answer:

Question 11.

Give a reason why concentrated and not dil H₂SO₄ – behave as an oxidising and dehydrating agent.
Answer:

1. Cone. H₂SO₄ be aves as an oxidising agent because cone.H₂SO₄ whes h. ated decomposes to give nascent oxygen which acts as a strong oxidising agent.On the other hand, dil H₂SO₄ on heating does not decompose to give nascent oxygen and as such cannot behave as an oxidising agent.

2. Cone. H₂SO₄ behaves as a dehydrating agent because cone. H₂SO₄ act as a great affinity for water and hydration of cone. H₂SO₄ is an exothermic reaction.On the other hand, dil H₂SO₄ has no affinity for water and hence cannot act as a hydrating agent.

Question 12.

Give the equation for the reaction cone, sulphuric acid with –

1.  glucose
2. sucrose
3. cellulose
4. an organic acid containing one carbon atom and two hydrogen atoms
5. an organic acid containing two carbon and two hydrogen atoms
6. an alcohol
7. hydrated copper (II) sulphate.

Answer:

Question 13.

State the observation seen when cone. H₂SO₄ is added to –
(1) sucrose
(2 ) hydrated copper (II) sulphate.
Answer:

1. Cone. H₂SO₄ dehydrates sucrose to carbon, called sugar charcoal. This is m the form of a black spongy charged mass of carbon.

2. Cone. H₂S0₄ dehydrates blue crystals of copper (II) sulphate pentahydrate (blue vitriol) to copper sulphite, which is in the form of a white powder.

Question 14.

State how addition of –

1. copper
2. NaCl to hot cone. H₂SO₄ serves as a test for the latter.

Answer:

1.Copper turnings when heated with cone. H₂SO₄ gives a colourless suffocating gas with a smell of burning sulphur(SO₂). The gas turns orange coloured acidified potassium dichromate solution green. This can be used as a test for cone. H₂SO₄.

2.Common salt (NaCl) when heated with cone. H₂SO₄ gives a colourless gas pungent smell only which (HCl) gives white fumes with NH₃. This can be used as a test for cone. H₂SO₄.

Question 15.

Give two tests for dilute sulphuric acid with balanced equations. State why
(1) BaCl₂
(2)  Pb(NO₃)₂ are used for the above tests.
Answer:

1. Barrium chloride solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in all acids
   BaCl₂ + H₂SO₄ (dil) → 2HCl + BaS0₄ ↓
2. Lead nitrate on treating with dilute sulphuric acid forms white ppt of lead sulphate, which is insoluble in all acids.
   Pb (NO₃)₂ + H₂SO₄ (dil) → 2HNO₃ + PbSO₄ ↓

Question 16.

Give a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO₃.
Answer:
Test to distinguish dil H₂SO₄ from dil HCl and HNO_3- BaCl₂ soln. when added to dil. H₂S0₄ gives a white ppt. of BaSO₄, but with dil. HCl and dil. HNO₃, no white ppt. is produced – since BaCl₂ and Ba(NO₃)₂ are soluble in dil. H₂SO₄.

Question 17.

State three different chemical compounds other than acids manufactured industrially from sulphuric acid.
Answer:

1. Barium sulphate
2. Lead sulphate
3. Sodium sulphate

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