A 0.1 molal aqueous solution of a weak acid is 30% ionised. If Kf

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 Multiple Choice QuestionsMultiple Choice Questions

321.

To observe an elevation of boiling point of 0.05°C, the amount of a solute ( mol. wt. = 100) to be added to 100 g of water (Kb= 0.5) is

  • 2g

  • 0.5g

  • 1g

  • 0.75g


322.

The measured freezing point depression for a 0.1 m aqueous CH3COOH solution is 0.19°C. The acid dissociation constant Ka at this concentration will be (Given, Kf the molal cryoscopic constant = 1.86 K kg mol-1)

  • 4.76 × 10-5

  • 4 × 10-5

  • 8 × 10-5

  • 2 × 10-5


323.

58.4 g of NaCl and 180 g of glucose were separately dissolved in l000 mL of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

  • NaCl solution will show higher elevation of boiling point

  • Glucose solution will show higher elevation of boiling point

  • Both the solutions will show equal elevation of boiling point

  • The boiling point elevation will be shown by neither of the solutions


324.

Which of the following will show a negative deviation from Raoult's law?

  • Acetone-benzene

  • Acetone-ethanol

  • Benzene-methanol

  • Acetone-chloroform


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325.

A saturated solution of H2S in 0.1 M HCl at 25°C contains S2- ion concentration of 10-23 mol L-1. The solubility product of some sulphides are CuS = 10-44, FeS = 10-14, MnS = 10-15 , CdS = 10-25 . If 0.01 M solution of these salts in 1M HCl are saturated with H2S, which of these will be precipitated?

  • All

  • All except MnS

  • All except MnS and FeS

  • Only CuS


326.

If the elevation in boiling point of a solution of 10 g of solute (mol. wt. = 100) in 100 g of water is Tb the ebullioscopic constant of water is

  • 10

  • 100 Tb

  • Tb

  • Tb10


327.

Which of these ions is expected to be coloured in aqueous solution?

I. Fe3+

II. Ni2+

III. Al3+

  • I and II

  • II and III

  • I and III

  • I, II and III


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328.

A 0.1 molal aqueous solution of a weak acid is 30% ionised. If Kf for water is 1.86° C/m, the freezing point of the solution will be

  • -0.18°C

  • -0.54°C

  • -0.36°C

  • -0.24°C


D.

-0.24°C

Freezing point depression (Tf) = iKfmHA     H+ + A+1-α        α         α1-0.3     0.3      0.3i= 1-0.3+0.3+0.3i= 1.3Tf= 1.3 × 1.86 ×0.1 =0.2418°CTf= 0-0.2418°C    = -0.2418°C


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329.

The freezing point of a solution composed of 10.0 g of KCl in 100 g of water is 4.5°C. Calculate the van't Hoff factor, i for this solution.

  • 2.50

  • 1.8

  • 1.2

  • 1.3


330.

During the depression in freezing point experiment, an equilibrium is established between the molecules of

  • liquid solvent and solid solvent

  • liquid solute and solid solvent

  • liquid solute and solid solute

  • liquid solvent and solid solute


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