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

371.

Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be

  • 3.98 × 108

  • 3.88 × 106

  • 3.68 × 10-6

  • 3.98 × 10-6


372.

For the following equilibrium, N2O4 2NO2 in the gaseous phase, NO2 is 50% of the total volume when equilibrium is set up. Hence, per cent of dissociation of N2Ois

  • 50%

  • 25%

  • 66.66%

  • 33.33%


373.

Of the given anions, the strongest Bronsted base is

  • ClO-

  • ClO3-

  • ClO2-

  • ClO4-


374.

At equilibrium, if KP = 1, then

  • G° > 1

  • G° < 1

  • G° = 0

  • G° = 1


Advertisement
375.

Which of the following is most acidic?

  • H2O

  • H2S

  • H2Se

  • H2Te


376.

1 mL of 0.01 N HCl is added to 999 mL solution of 0.1 N Na2SO4. The pH of the resulting solution will

  • 2

  • 7

  • 5

  • 1


377.

The equilibrium constants for the reaction,

Br2  2Br → 1

at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is

  • endothermic

  • exothermic

  • fast

  • slow


378.

A chemist wishes to prepare a buffer solution of pH = 2.90 that efficiently resists a change in pH yet contains only small concentration of buffering agents which one of the following weak acid along with its salt would be best to use 

  • m-chlorobenzoic acid (pKa = 3.98)

  • Acetoacetic acid (pKa = 3.58)

  • 2 5-dihydrobenzoic acid (pKa = 2.97)

  • p-chlorocmanic acid (pKa = 4.41)


Advertisement
Advertisement

379.

NaOH is a strong base. What will the be pH of 5.0 × 10-2 M NaOH solution? (log 2 = 0.3)

  • 14.00

  • 13.70

  • 13.00

  • 12.70


D.

12.70

Given that,

5.0 × 10-2 M NaOH = [OH-] = 5 × 10-2 M

[H+] [OH-] = 1 × 10-14

[H+] 5 × 10-2 = 1 × 10-14

[H+] = 1 × 10-145 × 10-2 = 2 × 10-13

pH = -log [H+] = -log (2 × 10-13)

     = 12.69 ≈ 12.70


Advertisement
380.

The equilibrium constants of the reactions

SO2(g) + 12O2(g)  SO3(g) and 2SO2(g) + O2(g)  2SO3(g) are K1 respectively. The relationship between K1 and K2 will be

  • K12=K2

  • K2 = √K1

  • K1= K2

  • K23 = K1


Advertisement