CBSE Class 12 Chemistry Solutions Assignment

Read and download free pdf of CBSE Class 12 Chemistry Solutions Assignment. Get printable school Assignments for Class 12 Chemistry. Class 12 students should practise questions and answers given here for Unit 02 Solutions Chemistry in Class 12 which will help them to strengthen their understanding of all important topics. Students should also download free pdf of Printable Worksheets for Class 12 Chemistry prepared as per the latest books and syllabus issued by NCERT, CBSE, KVS and do problems daily to score better marks in tests and examinations

Assignment for Class 12 Chemistry Unit 02 Solutions

Class 12 Chemistry students should refer to the following printable assignment in Pdf for Unit 02 Solutions in Class 12. This test paper with questions and answers for Class 12 Chemistry will be very useful for exams and help you to score good marks

Unit 02 Solutions Class 12 Chemistry Assignment

LEVEL-1 QUESTIONS

2.1 What are solns?
Homogeneous mixture containing two or more than two components.

2.2 What are binary solns?
Binary solution : Solutions containing only two components.

2.3.Define mole fraction
Ratio of number of moles of a component to the total number of moles of solutions.

2.4 Define molarity
Number of moles of solute dissolved in one litre of solution.

2.5 Define molality
Number of moles of the solute dissolved in 1 kg of the solvent.

2.6. State Henry’s law
The partial pressure of the gas in vapour phase is directly proportional to the molefraction of the gas in the solution. P = KH x

2.7. Mention one application of Henry’s law
To increase the solubility of CO2 in soft drinks , the bottle is sealed under high pressure.

2.8. State Raoutlt’s Law
For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. P1 = p0 1 x1

2.9. What are Ideal solutions?
Solutions which obey Raoult’s law . solute –solvent interaction is nearly equal to solute –solute or solvent – solvent interaction. Δmix H = 0 and Δmix V = 0 Eg: n- hexane and n-heptane

2.10. What are non-ideal solutions?
Solutions which do not obey Raoult’s law . Inter molecular force of attraction between solute – solvent is Weaker or stronger than those between solute –solute or solvent – solvent interaction. Δmix H ≠ 0 Δmix ↑ ≠ 0

Positive deviation: Solutions which do not obey Raoult’s law. Inter molecular force of attraction
between solute –solvent is Weaker than those between solute –solute or solvent – solvent interaction. Molecule find it easier to escape than in pure state. This will increase the vapour pressure and results positive deviation. Δmix H = + ive Δmix V = + ive eg. Mixtureof ethanol and water.

 

LEVEL-1 QUESTIONS

2.1 What are solns?
Homogeneous mixture containing two or more than two components.
 
2.2 What are binary solns?
Binary solution : Solutions containing only two components.
 
2.3.Define mole fraction
Ratio of number of moles of a component to the total number of moles of solutions.
 
2.4 Define molarity
Number of moles of solute dissolved in one litre of solution.
 
2.5 Define molality
Number of moles of the solute dissolved in 1 kg of the solvent.
 
2.6. State Henry’s law
The partial pressure of the gas in vapour phase is directly proportional to the molefraction of the gas in the solution. P = KH x
 
2.7. Mention one application of Henry’s law
To increase the solubility of CO2 in soft drinks , the bottle is sealed under high pressure.
 
2.8. State Raoutlt’s Law
For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. P1 = p01 x1
 
Important Questions for NCERT Class 12 Chemistry Solutions
 

Question. A solution has a 1 : 4 mole ratio of pentane to hexane.
The vapour pressures of the pure hydrocarbons at 20 °C are 440 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be
(a) 0.200
(b) 0.549
(c) 0.786
(d) 0.478 

Answer : D

Question. A solution of sucrose (molar mass = 342 g mol–1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1)
(a) – 0.372 °C
(b) – 0.520 °C
(c) + 0.372 °C
(d) – 0.570 °C 

Answer : A

Question. A solution of urea (mol. mass 56 g mol–1) boils at 100.18°C at the atmospheric pressure. If Kf and Kb for water are 1.86 and 0.512 K kg mol–1 respectively, the above solution will freeze at
(a) 0.654°C
(b) – 0.654°C
(c) 6.54°C
(d) – 6.54°C

Answer : B

Question. Pure water can be obtained from sea water by
(a) centrifugation
(b) plasmolysis
(c) reverse osmosis
(d) sedimentation.

Answer : C

Question. From the colligative properties of solution, which one is the best method for the determination of molecular weight of proteins and polymers?
(a) Osmotic pressure
(b) Lowering in vapour pressure
(c) Lowering in freezing point
(d) Elevation in boiling point 

Answer : A

Question. The relationship between osmotic pressure at 273 K when 10 g glucose (p1), 10 g urea (p2), and 10 g sucrose (p3) are dissolved in 250 mL of water is
(a) p2 > p1 > p3
(b) p2 > p3 > p1
(c) p1 > p2 > p3
(d) p3 > p1 > p2 

Answer : A

Question. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to
(a) mole fraction of solute
(b) mole fraction of solvent
(c) moles of solute
(d) moles of solvent.

Answer : A

Question. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that
(a) there will be no net movement across the membrane
(b) glucose will flow towards glucose solution
(c) urea will flow towards glucose solution
(d) water will flow from urea solution to glucose.

Answer : A

Question. A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at –0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = –1.86 °C/m)
(a) 3
(b) 4
(c) 1
(d) 2

Answer : D

Question. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol–1, the lowering in freezing point of the solution is
(a) 0.56 K
(b) 1.12 K
(c) –0.56 K
(d) –1.12 K

Answer : B

Question. Which one is a colligative property?
(a) Boiling point
(b) Vapour pressure
(c) Osmotic pressure
(d) Freezing point 

Answer : C

Question. Blood cells retain their normal shape in solution which are
(a) hypotonic to blood
(b) isotonic to blood
(c) hypertonic to blood
(d) equinormal to blood.

Answer : B

Question. The relative lowering of the vapour pressure is equal to the ratio between the number of
(a) solute molecules to the solvent molecules
(b) solute molecules to the total molecules in the solution
(c) solvent molecules to the total molecules in the solution
(d) solvent molecules to the total number of ions of the solute.

Answer : B

Question. The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is
(a) 0
(b) 1
(c) 2
(d) 3

Answer : D

Question. Which of the following 0.10 m aqueous solution will have the lowest freezing point?
(a) KI
(b) C12H22O11
(c) Al2(SO4)3
(d) C5H10O5 

Answer : C

 

Please click the link below to download CBSE Class 12 Chemistry Solutions Assignment
Unit 12 Aldehydes, Ketones and Carboxylic Acids
CBSE Class 12 Chemistry Aldehydes Ketons Carboxylic Acids Questions

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CBSE Class 12 Chemistry Unit 02 Solutions Assignment

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Assignment for Chemistry CBSE Class 12 Unit 02 Solutions

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Unit 02 Solutions Assignment Chemistry CBSE Class 12

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Unit 02 Solutions Assignment CBSE Class 12 Chemistry

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CBSE Chemistry Class 12 Unit 02 Solutions Assignment

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