Tamil Nadu Board of Secondary EducationHSC Science Class 11th

Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 - Solutions [Latest edition]

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Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 - Solutions - Shaalaa.com
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Chapter 9: Solutions

Evaluation
Evaluation [Pages 61 - 64]

Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide Chapter 9 Solutions Evaluation [Pages 61 - 64]

Choose the best answer

Evaluation | Q I. 1. | Page 61

The molality of a solution containing 1.8 g of glucose dissolved in 250 g of water is

  • 0.2 M

  • 0.01 M

  • 0.02 M

  • 0.04 M

Evaluation | Q I. 2. | Page 61

Which of the following concentration terms is/are independent of temperature

  • molality

  • molarity

  • mole fraction

  • molality and molarity

Evaluation | Q I. 3. | Page 61

Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide \[\ce{Al(OH)3 + 3HCl (aq) -> AlCl3 + 3H2O}\]

How many milliliters of 0.1 M Al(OH)3 solution is needed to neutralize 21 ml of 0.1 M HCl?

  • 14 mL

  • 7 mL

  • 21 mL

  • none of these

Evaluation | Q I. 4. | Page 61

The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 × 104 atm at 300 K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K?

  • 1 × 10-4

  • 1 × 104

  • 2 × 10-5

  • 1 × 10-5

Evaluation | Q I. 5. | Page 61

Henry’s law constant for the solubility of Nitrogen gas in water at 350 K is 8 × 104 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is

  • 4 × 10-4

  • 4 × 104

  • 2 × 10-2

  • 2.5 × 10-4

Evaluation | Q I. 6. | Page 61

Which one of the following is incorrect for an ideal solution?

  • ∆Hmix = 0

  • ∆Vmix = 0

  • ∆P = `"P"_"observed" - "P"_"calculated by Raoults law" = 0`

  • ∆Gmix = 0

Evaluation | Q I. 7. | Page 61

Which one of the following gases has the lowest value of Henry’s law constant?

  • N2

  • He

  • CO2

  • H2

Evaluation | Q I. 8. | Page 62

P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be

  • P1 + x1 (P2 – P1)

  • P2 – x1 (P2 + P1)

  • P1 – x2(P1 – P2)

  • P1 + x2(P1 – P2)

Evaluation | Q I. 9. | Page 62

Osomotic pressure (π) of a solution is given by the relation

  • π = nRT

  • πV = nRT

  • πRT = n

  • none of these

Evaluation | Q 10. | Page 62

Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?

  • Acetone + chloroform

  • Water + nitric acid

  • HCl + water

  • ethanol + water

Evaluation | Q I. 11. | Page 62

The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be

  • `(2x)/"y"`

  • `"y"/(0.2 x)`

  • `(0.2 x)/"y"`

  • `(5x)/"y"`

Evaluation | Q I. 12. | Page 62

At 100°C the vapour pressure of a solution containing 6.5 g a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be:

  • 102°C

  • 100°C

  • 101°C

  • 100.52°C

Evaluation | Q I. 13. | Page 62

According to Raoults law, the relative lowering of vapour pressure for a solution is equal to ______.

  • mole fraction of solvent

  • mole fraction of solute

  • number of moles of solute

  • number of moles of solvent

Evaluation | Q I. 14. | Page 62

At same temperature, which pair of the following solutions are isotonic?

  • 0.2 M BaCl2 and 0.2 M urea

  • 0.1 M glucose and 0.2 M urea

  • 0.1 M NaCl and 0.1 M K2SO4

  • 0.1 M Ba(NO3)2 and 0.1 M Na2SO4

Evaluation | Q I. 15. | Page 62

The empirical formula of a nonelectrolyte(X) is CH2 O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025M glucose solution at the same temperature. The molecular formula of X is

  • C2H4O2

  • C8H16O8

  • C4H8O4

  • CH2O

Evaluation | Q I. 16. | Page 62

The KH for the solution of oxygen dissolved in water is 4 × 104 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is ______.

  • 4.6 × 103

  • 1.6 × 104

  • 1 × 10-5

  • 1 × 105

Evaluation | Q I. 17. | Page 62

Normality of 1.25M sulphuric acid is ______.

  • 1.25 N

  • 3.75 N

  • 2.5 N

  • 2.25 N

Evaluation | Q I. 18. | Page 62

Two liquids X and Y on mixing gives a warm solution. The solution is ______.

  • ideal

  • non-ideal and shows positive deviation from Raoults law

  • ideal and shows negative deviation from Raoults Law

  • non-ideal and shows negative deviation from Raoults Law

Evaluation | Q I. 19. | Page 63

The relative lowering of vapour pressure of a sugar solution in water is 2.5 × 10-3. The mole fraction of water in that solution is ______.

  • 0.0035

  • 0.35

  • 0.0035/18

  • 0.9965

Evaluation | Q I. 20. | Page 63

The mass of a non – volatile solute (molar mass 80 g mol-1) should be dissolved in 92g of toluene to reduce its vapour pressure to 90% ______.

  • 10 g

  • 20 g

  • 9.2 g

  • 8 g

Evaluation | Q I. 21. | Page 63

For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is ______.

  • 310 × 0.082 K

  • 310° C

  • 37°C

  • `310/0.082`K

Evaluation | Q I. 22. | Page 63

200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be 2.52 × 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1)

  • 62.22 kg mol-1

  • 12444 g mol-1

  • 300 g mol-1

  • None of these

Evaluation | Q I. 23. | Page 63

The Van't Hoff factor (i) for a dilute aqueous solution of the strong elecrolyte barium hydroxide is (NEET) ______.

  • 0

  • 1

  • 2

  • 3

Evaluation | Q I. 24. | Page 63

Which is the molality of a 10% w/w aqueous sodium hydroxide solution?

  • 2.778

  • 2.5

  • 10

  • 0.4

Evaluation | Q I. 25. | Page 63

The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is

  • α = `("n"("i" - 1))/("n - 1")`

  • α2 = `("n"(1 - "i"))/("n - 1")`

  • α = `("n"("i" - 1))/("1 - n")`

  • α = `("n"(1 - "i"))/("n"("1 - i"))`

Evaluation | Q I. 26. | Page 63

Which of the following aqueous solutions has the highest boiling point?

  • 0.1 M KNO3

  • 0.1 M Na3PO4

  • 0.1 M BaCl2

  • 0.1 M K2SO4

Evaluation | Q I. 27. | Page 63

The freezing point depression constant for water is 1.86° K Kg mol-1. If 5 g Na2SO4 is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na2SO4 is ______.

  • 2.57

  • 2.63

  • 3.64

  • 5.50

Evaluation | Q I. 28. | Page 63

Equimolal aqueous solutions of NaCl and KCl are prepared, If the freezing point of NaCl is -2°C, the freezing point of KCl solution is expected to be ______.

  • -2°C

  • -4°C

  • -1°C

  • 0°C

Evaluation | Q I. 29. | Page 63

Phenol dimerizes in benzene having van’t Hoff factor 0.54. What is the degree of association?

  • 0.46

  • 92

  • 46

  • 0.92

Evaluation | Q I. 30. | Page 64

Assertion: An ideal solution obeys Raoults Law.

Reason: In an ideal solution, solvent – solvent as well as solute – solute interactions are similar to solute-solvent interactions.

  • Both assertion and reason are true and reason is the correct explanation of assertion

  • Both assertion and reason are true but reason is not the correct explanation of assertion.

  • Assertion is true but reason is false

  • Both assertion and reason are false

Write brief answer to the following questions.

Evaluation | Q II. 1. (i) | Page 64

Define molality.

Evaluation | Q II. 1. (ii) | Page 64

Define Normality.

Evaluation | Q II. 2. | Page 64

What is a vapour pressure of liquid?

Evaluation | Q II. 2. | Page 64

What is a relative lowering of vapour pressure?

Evaluation | Q II. 3. | Page 64

State and explain Henry’s law.

Evaluation | Q II. 4. | Page 64

State Raoult law and obtain the expression for lowering of vapour pressure when the nonvolatile solute is dissolved Insolvent.

Evaluation | Q II. 5. | Page 64

What is molal depression constant? Does it depend on nature of the solute?

Evaluation | Q II. 6. | Page 64

What is osmosis?

Evaluation | Q II. 7. | Page 64

Define isotonic solutions

Evaluation | Q II. 8. | Page 64

You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one super saturated. How would you determine each solution?

Evaluation | Q II. 9. | Page 64

Explain the effect of pressure on solubility.

Evaluation | Q II. 10. | Page 64

A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL-1 calculate the molality.

Evaluation | Q II. 11. | Page 64

A 0.25 M glucose solution, at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?

Evaluation | Q II. 12. | Page 64

Calculate the molality of a solution containing 7.5 g glycine(NH2-CH2-COOH) dissolved in 500g of water.

Evaluation | Q II. 13. | Page 64

Which solution has the lower freezing point? 10 g of methanol (CH3OH) in 100g g of water (or) 20 g of ethanol (C2H5OH) in 200 g of water.

Evaluation | Q II. 14. | Page 64

How many moles of solute particles are present in one liter of 10-4 M potassium sulphate?

Evaluation | Q II. 15. | Page 64

Henry’s law constant for solubility of methane in benzene is 4.2 × 10-5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at

  1. 750 mm Hg
  2. 840 mm Hg.
Evaluation | Q II. 16. | Page 64

The observed depression in the freezing point of water for a particular solution is 0.093°C calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K Kg mol-1.

Evaluation | Q II. 17. | Page 64

The vapour pressure of pure benzene (C6H6) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?

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Chapter 9: Solutions

Evaluation
Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 - Solutions - Shaalaa.com

Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 - Solutions

Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 (Solutions) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Tamil Nadu Board of Secondary Education Class 11th Chemistry Volume 1 and 2 Answers Guide solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Class 11th Chemistry Volume 1 and 2 Answers Guide chapter 9 Solutions are Solutions, Types of Solutions, Expressing Concentration of Solutions, Solubility of the Solutes, Henry's Law, Vapour Pressure of Liquid, Vapour Pressure of Liquid Solutions - Introduction, Ideal and Non-ideal Solutions, Colligative Properties, Reverse Osmosis (RO), Abnormal Molar Masses.

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