Skip to main content

Carbon and its Compounds class 10 CBSE | NCERT Solutions

NCERT Solution, MCQs, Study Notes, Q & A

1.Define the term, functional group and give two examples for it. 

Ans:
An atom or a group of atoms on which properties of the carbon compound depends is known as functional group. Eg: -
OH-            Alcoholic
- CHO        Aldehydic
- CO           Ketonic
 - COOH     Carboxylic

2. Name the functional groups present in the following compounds.

CH3CH2CH2-OH;  CH3CH2CH2COOH ; CH3-CH2-CHO;CH3COCH2CH2CH3
 Ans:
CH3CH2CH2-OH          Alcoholic
 CH3CH2CH2COOH     Carboxylic
CH3-CH2-CHO              Aldehydic
CH3COCH2CH2CH3      Ketonic

3. Describe the preparation of ethanol by the fermentation process. 

Ans:Fermentation of sugar is a process in which the sugar molecules are broken down in the ethylalcohol and carbon dioxide by the action of enzymes called invertage and zymase which are present in yeast. Molasses the bi-product of the sugar industry is mixed with suitable quantity of water and yeast powder. This mixer is kept in a huge steel tank at 250C to 300C for about 10 to 12 days, during this time fermentation takes place to form C2H5OH and CO2 gas. C2H5OH is recovered from this mixture by distillation.
                                   Invertase
C12 H22 O11 + H2O  -------------->        C6H12O6           +          C6H12O6
                               (From yeast)            Glucose                          fructose
           Zymase
C6 H22 O16                                       ----------------->  2C2H5OH + 2CO2
Glucose & Fructose                   (from yeast)                        Ethanol

4. What is meant by denaturated alcohol ? How is it prepared?

Ans:
Ethyl alcohol (ethanol) made into unfit for drinking is called denaturated alcohol.
It is prepared by adding poisonous substances like menthol, pyridine,  copper sulphate etc., to ethanol.

5. Give the names of the following :
i) an aldebyde derived from ethane
 ii) A ketone derived from butane
iii) The compound obtained by oxidation of ethanol by chromic anhydride (CrO3); and
iv) The substance formed on catalytic hydrogenation of methanal

Ans
(i) Ethanal (Or)Acetaldehyde CH3CHO
( ii) 2-Butanone (Or) Butanone-2 CH3COCH2CH3
(iii) Ethanal (Or)Acetaldehyde CH3CHO
(iv) Methane CH4

5. How is ethanoic acid prepared commercially from methanol ?

Ans:  Ethanoic acid is prepared commercially by reaction between methanol and carbon monoxide in the presence of Iodine - rhodium catalyst.
                                     I2-Rh
CH3OH + CO   ------------------> CH3COOH

6. Explain the follwing terms: (i)Esterification (ii)saponification (iii) decarbodylation

Ans: Esterification:- The reaction of carboxylic acid with an alcohol to form ester in the presence of conc. H2SO4 know as Esterifilation reaction.
CH3COOH + C2H5OH --------------à CH3COOC2H5 + H2O
Saponification : oil or a fat on heating with sodium hydroxide (NaOH) Solution forms sodium salt of carboxylic acid (soap)and glycerol, this reaction is know as saponification reaction.
CH2 – O COR                         CH2OH
|                                                  |
CH – O COR + 3NaOH       CHOH + 3RCOO Na
|                                                  |
CH2 – O COR       --------->        CH2OH

Decarboxylation: Sodium salt of ethanoic acid and soda lime on heating forms methane gas. This reaction is known as Decarboxylation reaction.
CH3COONa + NaOH         ------------->      CH4 + Na2CO3

7. Write chemical equations of the reactions of ethanoic acid with i)sodium (ii) sodium carbonates (iii) ethanol in presence of conc. H2SO4 and (iv) soda lime

Ans: (i) 2 CH3COOH + 2Na ------------------>2CH3COONa + H2­
 (ii) 2CH3COOH + Na2CO3  --------------------------->  2CH3COONa + CO2 + H2O
(iii) CH3COOH + C2H5OH  -------------------------->     CH3COOC2H5 + H2O
(iv) CH3COOH + NaOH   ---------  ----------------------> CH4 + Na2CO3


8. “Saturated hydrocarbons burn with a blue flame while unsaturated hydrocarbons burn with a sooty flame”. Why?

Ans: It is because a saturated hydrocarbon undergoes complete combustion due to high amount of carbon.
C2H6 (CNG) + 7O2 ---------------> 4CO2 + 6H2O + Heat

9. How ethanoic acid got its name as glacial acetic acid?

Ans: Ethanoic acid  has melting point is 290K so it often freezes during winter and looks like glacier. Therefore, it is also known as glacial acetic acid.

10. Name the products obtained on complete combustion of ethanol.

Ans: Carbon dioxide water and Heat

View NCERT solution
NCERT Solutions

Links For   CBSE Class X Science Term-II
For Solved CBSE Exam questions Visit  Read more

Comments

CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get dissolved in it

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm. Therefor

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page

Living science ratna sagar class 6 solutions

Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II

Electricity numerical for class 10 CBSE Trend Setter 50 Problems

1. The current passing through a room heater has been halved. What will happen to the heat produced by it? 2. An electric iron of resistance 20 ohm draws a current of 5 amperes. Calculate the heat produced in 30 seconds. 3. An electric heater of resistance 8 ohm takes a current of 15 A from the mains supply line. Calculate the rate at which heat is developed in the heater. 4. A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination in half a minute. 5. A resistance of 25 ohm is connected to a 12 V battery. Calculate the heat energy in joules generated per minute. 6. 100 joules of heat is produced per second in a 4 ohm resistor. What is the potential difference across the resistor? 7. An electric iron is connected to the mains power supply of 220 V. When the electric iron is adjusted at minimum heating’ it consumes a power of 360 W but at ‘maximum heating’ it takes a power of 840 W. Ca