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CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-8

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The one subject in NEET which is candidates who can easily attain good marks is Chemistry. That's the reason, often, one doesn’t pay notice and choose to compromise it. But if one wants to rank above others, the tip is to be thorough with NEET chemistry concepts. The understanding of reactions and definite basic understanding is what requires major attention in Chemistry but once done it only gets simpler from there. The main focus on the to-do list should be on getting a hang of the NCERT syllabus of NEET chemistry.
.

Q1. pH value of which one of the following is not equal to one?
  •   0.1 M HNO3
  •  0.05 M H2SO4
  •  0.1 M CH3COOH
  •  50 cm3 of 0.4 M HCl+50 cm3 of 0.2 M NaOH
Solution
(c) Among the given, pH of 0.1M CH3COOH is not equal to one as CH3COOH is a weak acid, thus does not ionise completely.

Q2.Which acts both as Lowry Bronsted acid and base?
  •  OH-
  •  Na2CO3
  •  NH3
  •  HSO4-
Solution
(d) HSO4- can accept a proton or can donate a proton (forms SO4(2-)).
Q3.  In the hydrolytic equilibrium, A-+H2O⇌HA+OH-
Ka=1.0×10-5.The degree of hydrolysis of 0.001 M solution of the salt is:
  •  10-3
  •  10-4
  •  10-5
  •  10-6
Solution
a)

Q4. The equilibrium constant (Kc) for the reaction, N2(g)+O2(g) ⇌2NO(g) at room temperature T is 4 × 10-4. The value of Kc for NO(g)⇌1/2 N2(g)+1/2 O2(g) at the same T is :
  •  0.02
  •  50
  •  4 × 10-4
  •  2.5×10-2
Solution
b)

Q5.The correct order of increasing [H3O+] in the following aqueous solutions is:
  •  0.01 M H2S<0.01 M H2SO4<0.01 M NaCl<0.01 M NaNO2
  •  0.01 M NaCl<0.01 M NaNO2<0.01 M H2S<0.01 M H2SO4
  •  0.01 M NaNO2<0.01 M NaCl<0.01 M H2S<0.01 M H2SO4
  •   0.01 M H2S<0.01 M NaNO2<0.01 M NaCl<0.01 M H2SO4
Solution
 c) H2SO4 is strong acid having pH<7. NaNO2 on hydrolysis gives alkaline solution of pH>7. NaCl is neutral and H2S is weak acid

Q6.In the reaction, AlCl3+Cl-⟶[AlCl4]-,AlCl3 acts as:
  •  Salt
  •  Lewis base
  •  Lewis acid
  •  Bronsted acid
Solution
c) AlCl3 accepts electron pair.

Q7.On doubling p and V with constant temperature, the equilibrium constant will
  •   Remain constant
  •  Become double
  •  Become one-fourth
  •  None of these
Solution
(a) At constant temperature pV constant On doubling p and V with constant T, the equilibrium constant (K) will remain constant. Pressure will never affect the value of K. It may result in the shifting of equilibrium, but not the equilibrium constant. By doubling the volume, the concentration of both reactants and products evenly became half. Therefore, over all there is no change in equilibrium constant value (K).

Q8.Hydroxyl ion concentration of 10-2 M HCl is
  •  1×101 mol dm-3
  •  1×10-12 mol dm-3
  •  1×10-1 mol dm-3
  •  1×10-14 mol dm-3
Solution
b)

Q9.The solubility of CaF2 in pure water is 2.3×10-6 mol dm-3.Its solubility product will be
  •   4.8×10-18
  •  48.66×10-18
  •  4.9×10-11
  •  48.66×10-15
Solution
b) CaF2⇌Ca(2+)+2F- s 2s Ksp=s(2s)2=4s3 Ksp=4(2.3×10(-6) )3

Q10. The number of mole of hydroxide [OH-] ion in 0.3 litre of 0.005 M solution of Ba(OH)2 is:
  •  0.0075
  •  0.0015
  •  0.0030
  •  0.0050
Solution
c) Mole OH-=M×V(in litre) ∴No of OH-=0.3×0.005×2=0.0030

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