Would gaseous HCl be considered as an Arrhenius acid ? from

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681.

A plot of 1T vs k for a reaction gives the slope -1 × 104 K. The energy of activation for the reaction is (Given, R = 8.314 K-1 mol-1)

  • 8314 J mol-1

  • 1.202 kJ mol-1

  • 12.02 J mol-1

  • 83.14 kJ mol-1


682.

One mole of ammonia was completely absorbed in one litre solution each of

I. 1 M HCl

II. 1 M CH3COOH 

III. 1 M H2SO4 at 298 K

The decreasing order for the pH of the resulting solution is (Given Kb (NH3) = 4.74)

  • II > III > I

  • I > II > III

  • II > I > III

  • III > II > I


683.

Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1.85 10-6 Sm-1. Solubility product of the salt AB at 298 K is

(Given, °m (AB) = 140  Sm2 mol-1)

  • 5.7 × 10-12

  • 1.32 × 10-12

  • 7.5 × 10-12

  • 1.74 


684.

For the properties mentioned, the correct trend for the different species is in

  • strength as Lewis acid - BCl3 > AlCl3 > GaCl3

  • inert pair effect - Al > Ga > ln

  • oxidising property - Al3+ > ln3+ > Ti3+

  • first ionization enthalpy - B > Al > Ti


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685.

In the reaction,

Fe(OH)3 (s)  Fe3+ (aq) + 3OH- (aq), if the concentration of OH- ions is decreased by 14 times, then the equilibrium concentration of Fe3+ will increase by

  • 8 times

  • 16 times

  • 64 times

  • 4 times


686.

Equilibrium constants K1 and K2 for the following equilibria,

I. NO (g) + 12 O2 (g)  NO2 (g)

II. 2NO2 (g)  2NO (g) + O2 (g) are related as

  • K1K2

  • K21K1

  • K1 = 2K2

  • K21K12


687.

The pair of compound which cannot exist together in solution is

  • NaHCO3 and NaOH

  • NaHCO3 and H2O

  • NaHCO3 and Na2CO3

  • Na2COand NaOH


688.

Which is most basic in character ?

  • CsOH

  • KOH

  • NaOH

  • LiOH


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689.

Would gaseous HCl be considered as an Arrhenius acid ?

  • Yes

  • No

  • Not known

  • Gaseous HCl does not exist


B.

No

Gaseous HCl would not be considered as an Arrhenius acid.


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690.

The degree of hydrolysis in hydrolytic equilibrium

A- + H2 HZ + OH-

at salt concentration of 0.001 M is :

(Ka = 1 × 10-5)

  • 1 × 10-3

  • 1 × 10-4

  • 5 × 10-4

  • 1 × 10-6


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