Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

A vessel of one litre capacity containing 1 mole of SO3 is heated till a state of equilibrium is attained.

2SO3 (g)  2SO2 (g) + O2 (g)

At equilibrium, 0.6 moles of SO2 had formed. The value of equilibrium constant is

  • 0.18

  • 0.36

  • 0.45

  • 0.68


D.

0.68

Initial concentrationEquilibrium 2SO3 (g)1.0 mole0.4 mole  2SO2 (g)0.0 mole0.6 mole + O2 (g)0.0 mole0.3 mole

Equilibrium constant is given by

K = SO22 O2SO32 = 0.62 0.30.42 = 0.675  0.68


2.

The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion (α) of a gas, i.e., α = 1VdVdTP. For an ideal gas α is equal to

  • T

  • 1T

  • P

  • 1P


B.

1T

For n moles of an ideal gas PV = nRT

or VnRTP

Differentiating  with respect to at constant P, we have

dVdTP = nRP = VTα = 1VdVdTP = 1V × VT = 1T

Therefore, for an ideal gas α is equal to 1T.


3.

KF combines with HF to form KHF2. This compound contains the species

  • K+, F- and H+

  • K+, F- and HF

  • K+ and [HF2]-

  • [KHF]+ and F2


C.

K+ and [HF2]-

The fluoride ion, by a process of hydrogen bonding, forms the anion, HF2-. The compound is written as K+[HF2]-.


4.

The molecules having the same hybridization, shape and number of lone pairs of electrons are

  • SeF4, XeO2F2

  • SF4, XeF2

  • XeOF4, TeF4

  • SeCl4, XeF4


A.

SeF4, XeO2F2

SeF4 and XeO2F2 are both sp3 hybridized, trigonal bipyramidal and see-saw shaped with 1 lone pair of electrons each.

SF4 has 1 lone pair, XeF2 has 3 lone pairs.

XeOF4 is square pyramidal with 1 lone pair. TeF4 is see-saw shaped with 1 lone pair.

SeCl4 has 1 lone pair and is see-saw shaped. XeF4 has 2 lone pairs and is planar in shape.


5.

2.5 g of the carbonate of a metal was treated with 100 mL of 1N H2SO4. After the completion of the reaction, the solution was boiled off to expel CO2 and was then titrated against 1N NaOH solution. The volume of alkali that would be consumed, if the equivalent weight of the metal is 20

  • 50

  • 25

  • 75

  • 100


A.

50

Equivalent weight of metal carbonate = 20 + 30 = 50

2.5 gm of metal carbonate = 2.550 = 0.05 eq.

Number of equivalent of H2SO4 would have reacted = 0.05

Number of equivalent of H2SO4 taken = 100 × 11000= 0.1

Number of equivalent weight of H2SO4 reamins unreacted = 0.1 - 0.05 = 0.05 eq.

 Number of equivalent of alkali consumed = 0.05 eq.

milli eq = Normality × Volume in mL

1.0 × V = 0.05 × 1000

V = 0.05 × 10001.0 = 50 mL


6.

Mercurous chloride exists in the form of

  • Hg+

  • Hg22+

  • Hg2+

  • Hg32+


B.

Hg22+

Mercurous chloride or calomel mineral is a rare mineral. It is dense white or yellowisg- white, odorless solid. It is a component of reference electrodes in electrochemistry. It exists in the form of Hg22+.


7.

The M - O- M bond angles in MO (where M is halogen) is in the order

  • Br2O > Cl2O > F2O

  • F2O > Br2O > Cl2O

  • F2O > Cl2O > Br2O

  • Cl2O > F2O > Br2O


A.

Br2O > Cl2O > F2O

In OF2 repulsion between lone pairs is greater than that between bond pair since electrons are away from O and nearer to F. In Cl2O, bonding electrons are nearer to O than to Cl, so the bond angle is greater than 109°28'. In Br2O, the bonding electrons are more closer to oxygen than in Cl2O, so the bond angle is largest (116°).


8.

The equilibrium constant for mutarotation α-D Glucose  β-D Glucose is 1.8. What percentage of α form remains at equilibrium?

  • 35.7

  • 64.3

  • 55.6

  • 44.4


A.

35.7

InitialAt equilibrium α-D glucose11 - α  β-D glucose0αK = α1 - α

Solving α = 0.642

1 - α = 0.358

Percent of α-D glucose remaining at equilibrium = 35.8%


9.

What is the molarity of H2SO4 solution that has a density of 1.84 g/cc at 35°C and contains 98% by weight?

  • 4.18 M

  • 8.14 M

  • 18.4 M

  • 18 M


C.

18.4 M

Molarity = weight of solutemolecular weight × 1000volume of solution (in mL)

(Since, Volume of solution = massdensity)

Therefore, volume of solution = 1001.84 = 54.34 mL

Molarity = 9898 × 100054.34 = 18.4 M


10.

pH of a 0.01 M solution (Ka = 6.6 × 10-4)

  • 7.6

  • 8

  • 2.6

  • 5


C.

2.6

pH of a 0.01 M solution is 

α = KaC = 6.6 × 10-40.01 = 6.6 × 10-2 = 0.257H+  =  = 0.01 × 0.257 = 2.57 × 10-3

pH = 3 - log2.5 = 2.60