CHEMICAL EQUILIBRIUM QUIZ-8

CHEMISTRY CHEMICAL EQUILBRIUM QUIZ-8

Dear Readers,

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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^-
K_a=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 (K_c) for the reaction, N₂(g)+O₂(g) ⇌2NO(g) at room temperature T is 4 × 10^-4. The value of K_c 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 K_sp=s(2s)²=4s³ K_sp=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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