Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

ForthereactionSO2+12O2SO3,ifwewriteKp=Kc(RT)x,thenxbecome

  • -1

  • -12

  • 12

  • 1


B.

-12

Intherelation,Kp=Kc(RT)x,x=np-nrForthereaction,SO2(g)+12O2(g)SO3(g)x=1-1+12=-12


2.

Hybridisation of central atom in NF3 is

  • sp3

  • sp

  • sp2

  • dsp2


A.

sp3

Number of hybrid orbitals = number of σbonds + number of lone pairs

3σbonds + 1 lone pair = 4

.:. Hybridisation of N is sp3 and geometry of the molecule is pyramidal.


3.

For the Paschen series the values of n1and n2in the expressionE=Rhc1n12-1n22are

  • n1= 1, n2 = 2, 3, 4, ...

  • n1= 2, n2 = 3, 4, 5,...

  • n1= 3, n= 4,5,6,...

  • n1= 4, n2 = 5,6,7,...


C.

n1= 3, n= 4,5,6,...

Paschen series is the third series of hydrogen spectrum and lies in the infrared region, therefore for this series, n1 = 3 and n2> n1,ie, 4, 5, 6,....


4.

Under which of the following conditions is the relation H=E+PVvalid for a closed system ?

  • Constant pressure

  • Constant temperature

  • Constant temperature and pressure

  • Constant temperature, pressure and composition


A.

Constant pressure

At constant pressure,

p=0H=E+pVisvalid.


5.

The second law of thermodynamic says that in a cyclic process

  • work cannot be converted into heat

  • heat cannot be converted into work

  • work cannot be completely converted into heat

  • heat cannot be completely converted into work


D.

heat cannot be completely converted into work

Second law of thermodynamics states that in a cyclic process heat cannot be converted completely into work because attainment of 0 K temperature is impossible.


6.

The equilibrium constant (K) of a reaction may be written as

  • K = e-G/RT

  • K =e-G°/RT

  • K = e-H/RT

  • K = e-H°/RT


B.

K =e-G°/RT

Gibbsfreeenergy(G°)isrelatedwithequilibriumconstant(K)asG°=-RTlnK-G°RT=lnKK=e-G°RT


7.

1 mole of photon, each of frequency 2500 s-1, would have approximately a total energy of

  • 10 erg

  • 1 J

  • 1 eV

  • 1 MeV


A.

10 erg

Frequency of one photon = 2500 s-1
:. Frequency of 1 mole of photon
= 6.023 x 1023 x 2500 s-1

Totalenergyof1molofphoton,E==6.626×10-34Js×6.023×1023×2500s-1=9.97×10-7J10erg


8.

An organic compound made of C, H and N contains 20% nitrogen. Its molecular weight is

  • 70

  • 140

  • 100

  • 65


A.

70

Atomic mass of nitrogen= 14
Since the compound contains C, H and N, at
least one N atom must be there in the molecule.

20%N=14100%=1420×100=70

:. The minimum molecular weight of the organic compound = 70.


9.

Equal volumes of molar hydrochloric acid 'and sulphuric acid are neutralized by dilute NaOH solution and x kcal and y kcal of heat are liberated respectively. Which of the following is true ?

  • x = y

  • x =y/2

  • x = 2y

  • none of these


B.

x =y/2

Heat of neutralisation= Heat of formation of water (1 mol).

HCl+NaClNaCl+H2O;H=-xorH+1mol+OH-1molH2O;H=xH2SO4+2NaOHNa2SO4+2H2O;H=-yor2H++2OH-2H2O;H=-yH++OH-H2O;H=-y2H=Hx=y2


10.

Which of the following will decrease the pH of a 50 mL solution of 0.01 M HCl?

  • Addition of 5 mL of 1 M HCl

  • Addition of 50 mL of 0.01 M HCl

  • Addition of 50 mL of 0.002 M HCl

  • Addition of Mg


B.

Addition of 50 mL of 0.01 M HCl

Millimoles of 50 mL of 0.01 M HCl
= 50 x 0.01 M = 0.5
pH of the solution =-log (0.5) = 0.3010

(a) On adding 5 mL of 1 M HCl,
Concentration of the resultant solution

=50×0.01+5×150+5=0.1

:. pH of the solution = -log (0.1) = 1

(b) On adding 50 mL or 0.01 M HCl,
Concentration of the resultant solution

=50x0.01+50x0.0150+50=0.01

:. pH of the solution = -log (0.01) = 2

(c) On adding 50 mL of 0.002 M HCl,
concentration of the resultant solution

=50×0.01+50x0.00250+50=6×10-3

:. pH of the solution = -log (6 x 10-3)= 2.22

(d) On adding Mg,

Mg+2HClMgCl2+2H+Since,pH1[H+]:.InthiscasepHdecreases.