Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsMultiple Choice Questions

261.

A liquid is in equilibrium with its vapour at it's boiling point. On the average, the molecules in two phases have equal

  • inter- molecular forces

  • potential energy

  • kinetic energy

  • total energy


262.

For the chemical reaction 3X(g)+ Y(g) X3Y (g), the amount of X3Y at equilibrium is affected by

  • temperature and pressure

  • temperature only

  • pressure only

  • temperature, pressure and catalyst


263.

The pair of compound which cannot exists together in solution is

  • NaHCO3 and NaOH

  • Na2CO3 and NHCO3

  • Na2CO3 and NaOH

  • NaHCO3 and NaCl


264.

In homogeneous catalysis

  • the reactants, catalyst and products are in the same phase.

  • the catalyst and reactants are in the same phase.

  • the catalyst and products are in the same phase.

  • the reactants and products are in the same phase.


Advertisement
265.

Which one of the following pair shows Buffer's solution?

  • NaCl + NaOH

  • CH3COONa + CH3COOH

  • CH3COOH + CH3COONH4

  • H2SO4 + CuSO4


266.

Which of the following chloride is water insoluble?

  • HCl

  • AgCl

  • Both 'a' and 'b'

  • None of these


267.

0.005 M acid solution has 5 pH. The percentage ionisation of acid is :

  • 0.8%

  • 0.6%

  • 0.4%

  • 0.2%


268.

CH3COOH is weaker acid than H2SO4.It is due to:

  • more ionization

  • less ionization

  • covalent bond

  • electrovalent bond


Advertisement
269.

In the reaction 3A + 2B 2C, the equilibrium constant Kc is given by

  • [3A] x[2B][C]

  • [A]3×[B][C]

  • [C]2[A]3×[B]2

  • [C][3A][2B]


Advertisement

270.

The solubility of Sb2S3 in water is 1.0 × 10-5 mol/L at 298K. What will be its solubility product?

  • 108 × 10-25

  • 1.0 × 10-25

  • 144 × 10-25

  • 126 × 10-24


A.

108 × 10-25

Sb2S3s mol/L  2Sb3+2s + 3S2-3s

Solubility product (Ksp) = [Sb3+][S2-]3

                                   = (2s)2(3s)3 = 108 s5

                                   = 108 × (1.0 × 10-5)5

                                   = 108 × 10-25


Advertisement
Advertisement