Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

4.0 g molecular PCl5 were heated in a vessel of 6 L volume. At equilibrium state , 50% of PCl5 is dissociated. The equilibrium constant for the reaction is :

  • 0.66 mol L-1

  • 0.33 mol L-1

  • 0.44 mol L-1

  • 0.88 mol L-1


B.

0.33 mol L-1

                                PCl5  PCl3 + Cl2

Moles before

dissociation               C            0       0

Moles after

dissociation               C(1 - α)  Cα      Cα

Kc[PCl3] [Cl2][PCl5]= C2α2C(1 - α) = 2(1 - α)

Given , C = 46 and α = 0.5     C = molelitre

Kc4 × (0.5)26 × (1 - 0.5) =0.33 mol L-1


2.

One mole of FeC2O4 is oxidised by x moles of Cr2O72- in acidic medium , x is:

  • 0.5

  • 1.0

  • 1.5

  • 2.0


A.

0.5

Cr2O72-+ 14H+ + 6e-  2Cr3+ + 7H2O

                   FeC2O4  Fe3+ + 2CO2 + 3e- ] x 2

________________________________________

Cr2O72-1 mole + 2FeC2O42 moles + 14H+  2Cr3+ + 2Fe3+ + 4CO2

                                                       + 7H2O

Thus, one mole of FeC2O4 is oxidised by 0.5 mole of

Cr2O72- in acidic medium.


3.

The number of σ bonds in P4O10 is:

  • 6

  • 16

  • 7

  • 17


B.

16

The number of σ bonds in P4O10 is 16. There are also four π bonds presents in P4O10 molecule.


4.

The decreasing nucleophilicity of the following is:

I. CH3S-              II. CH3O-

III. -OH               IV. EtO-

  • II > III > 1V > I

  • III > IV > I > II

  • IV > III > I > II

  • II < I < IV < III


C.

IV > III > I > II

Acidity is explained as

CH3SH > CH3OH > H2O > EtOH

Basicity : CH3S- < CH3O- < -OH < EtO-

When nucleophilic centre is different (CH3Sand CH3O-) and they are from the same group, nucleophilicity is opposite to basicity.

 Nucleophilicity EtO- > -OH > CH3S- > CH3O-
Hence, the answer is (c)


5.

For a chemical reaction, temperature is raised from 25° C to 35° C. The value of equilibrium constant, Kp is doubled . The heat of reaction (ΔH) for the reaction will be :

  • 13.834 kcal

  • 11.714 kcal

  • 10.726 kcal

  • 4.312 kcal


C.

10.726 kcal

Given that , T1 = 25°C = 298 K
T1 = 35°C = 308 K

KP2KP1 = 2R = 2calK-1 mol-1

From van't Hoff equation ,

2.303log KP2KP1 = ΔHR T2 - T1T1T2

2.303 log 2 = ΔH2308 - 298308 × 298

or , ΔH = 12726 cal = 10.726 kcal


6.

Consider the following pairs of ions:

I. Sc3+ and Ti4+        II. Mn2+ and Fe2+

III. Fe2+ and Co3+    IV. Cu+ and Zn2+

Among these pairs ofions, isoelectronic pairs would include:

  • II, III and IV

  • I, III and IV

  • I, II and IV

  • I, II and III


B.

I, III and IV

I. Electrons in so3+ = 18 and in Ti4+ = 18
II. Electrons in Mn2+ =23 and in Fe2+ = 24
III. Electrons in Fe2+ = 24 and in Co3+ =24
IV. Electrons in Cu+ = 28 and in Zn2+ = 28


7.

A litre solution containing NH4Cl and NH4OH has hydroxide ion concentration of 10-6 mol L-1. Which of the following hydroxide could be precipitated when the solution is added to 1L solution of 0.1M metal ions?

Ba(OH)2 (Ksp = 5 x 10-3)

Ni(OH)2 (Ksp = 1.6 x 10-16)

Mn(OH)2 (Ksp = 2 x 10-13

Fe(OH)2 (Ksp = 8 x 10-16)

  • I and IV

  • III and IV

  • II and IV

  • II , III and IV


C.

II and IV

When 1L each solution are mixed

[OH]- = 10-6 M       (buffer solution)

Mn+ = 0.05M

for Q = [0.5][10-6]2 = 5 x 10-14

Q > Ksp for Fe2+ and Ni2+

Thus , Fe(OH)2 and Ni(OH)2 are precipitate.


8.

Arrange the following (I to IV) in the order of increasing masses:

I. 0.5 mole of O3

II. 0.5 g atom of oxygen

III. 3.011 x 1023 mlecules of O2

IV. 5.6 l of CO2 at STP

  • II < IV < III < I

  • IV < III < I < II

  • II < I < IV < III

  • I < II < III < IV


A.

II < IV < III < I

I. 0.5 mole O3 = 24 g O3

II. 0.5 g atom of oxygen = 8 g O2

III. 3.011 × 10236.022 × 1023 × 32 = 16 g of O2

IV. 5.622.4 × 44 g CO2 = 11 g CO2


9.

Forty millilitre of CO was mixed 100 mL of O2 and the mixture was exploded. On cooling, the reaction mixture was shaken with KOH. What volume of gas is left?

  • 80 ml of O2

  • 60 ml of O2

  • 80 ml of CO2

  • 50 ml of CO2


A.

80 ml of O2

CO40ml + 12O220 ml   CO240 ml  KOH absorbed

volume of O2 left = (100-20) = 80 ml


10.

10 moles of ideal gas expand isothermally and reversibly from a pressure of 10 atrn to 1 atm at 300K. What is the largest mass which can lifted through a height of 100 meter?

  • 48.75 kg

  • 58.55 kg

  • 75.64 kg

  • 64.13 kg


B.

58.55 kg

W = - nRT ln p1p2

= -10 x 8.314 x 300 ln101

= -57441.42 J

W = - mgh

 m X 9.81 x 100 = 57441.42

m = 58.55kg