Subject

Chemistry

Class

NEET Class 12

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

 Multiple Choice QuestionsMultiple Choice Questions

11.

The standard molar heat of formation of ethane ,CO2 and waterare respectively - 21.1 , - 94.1and - 68. 3 kcal .The standard molar heat ofcombustion of ethane will be :

  • - 372 kcal

  • 162 kcal

  • - 240 kcal

  • 183.5 kcal


A.

- 372 kcal

Combustion equation for ethane is :

C2H6 (g) +72O2 (g)2CO2 (g) + 3H2O (g)H°= ?

H°= {ΣHf°products} - {ΣHf°reactants}

= {2Hf°(CO2) + 3Hf (H2O)} - {Hf°(C2H6) + 7/2Hf°(O2)}

{2 X (- 94.1) + 3 X (- 68.3)} - {(-21.1) + 7/2(O)}

={- 1888.2 - 204.9} - {-21.1}

= - 393.1 + 21.1

= - 372 kcal


12.

Which of the following is paramagnetic ?

  • N2

  • C2

  • N2+

  • O22-


C.

N2+

C2➔ the molecular orbital configuration (12) is

➔ C2 = [(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2py)2 (π2pz)2]

C2 molecule is diamagnetic as it does notcontain any unpaired electron

N2the molecular orbital configuration (14) is

N2 = [(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2py)2 (π2pz)2 (σ2px)2]

N2 molecule is diamagnetic because it does notcontain any unpaired electron .

N2+the molecular orbital configuration (13) is

➔ N2+ = [(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2py)2 (π2pz)2 (σ2px)1]

N2+molecule is paramagnetic because it containone unpaired electron .

O22-the molecular orbital configuration (18) is

O22- = [(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2px)2 (π2py)2 (π2pz)2 (π*2py)2(π*2pz)2]

It does not contain any unpaired electron so , it isdiamagnetic .


13.

Intermolecular hydrogen bonding is strongest in :

  • C2H5OH

  • HF

  • H2O

  • CH3COCH3


B.

HF

···Hδ+-Fδ-···Hδ+-Fδ-···

Strength of H · · · · · F > H · · · · · O


14.

NH4COONH2 (s)2NH3 (g) + CO2 (g) ,ifequlibrium pressure is 3 atm for the abovereaction .Kpfor the reaction is :

  • 4

  • 27

  • 4/27

  • 1/27


B.

27

NH2COONH4 (s)2NH3 (g) + CO2 (g)

On applying the law of mass action to thisheterogeneous system

Kp =PNH32×PCO2PNH2COONH4(s)

or Kp =PNH32XPCO2

= 32X3

= 9 X 3 = 27


15.

Non - directional orbital is :

  • 4p

  • 4d

  • 4f

  • 3s


D.

3s

Non - directional orbital is : 3s because s - sub - shellhas only one orbital and it is spherical .Hence ,s - orbitals are non - directional .


16.

Which of the following gas is linear ?

  • CO2

  • SO2

  • NO2

  • SO3


A.

CO2

In CO2 ,the bond length (C-O) is 1.15A°,hybridization of C in CO2 is sp .Thus it has alinear structure .

O=C=OO+=C-O-O--CO+ .


17.

Which of the following is most polarised ?

  • Kr

  • He

  • Ar

  • Xe


D.

Xe

Due to increase in the atomic size of zero groupelements from He to Xe , the polarizability oftheir atom by a polar solvent increases in thesame order .Thus Xe is most polarised .


18.

72X1802αβγzXA , Z and A are :

  • 69 , 172

  • 172 , 69

  • 180 , 70

  • 182 , 68


A.

69 , 172

When an alpha particle is ejected , results in

➔ decrease in atomic weight by four units

➔ decrease in atomic number of twoWhen a beta particle is ejected , results in

➔ no change in atomic weight

➔ increase in atomic number by one unitwhen anγ particle is ejected results in

➔ nochange

72X1802α68A172-1β69B172γ69X172


19.

A radioactive nucleus will not emit :

  • alpha and beta rays simultaneously

  • beta and gamma rays simultaneously

  • gamma and alpha rays

  • gamma rays only


D.

gamma rays only

When a radioactive nucleus are passed throughelectric or magnetic field , they are split intoradioactive rays .A radioactive substance firstemitsα andβ radiations , then it becomesunstable and emit rays .


20.

If the Vrms is 30 R1/2 at 27oCthen calculate themolar mass of gas in kilogram .

  • 1 Kg

  • 2 Kg

  • 0.001 Kg

  • 0.002 Kg


C.

0.001 Kg

We know ,

Vrms =3PVM

=3RTM

Given Vrms = 30R1/2

T = 273 + 27 = 300 K

30R1/2 =3RTM

302R=3RTM

30 x 30 x R =3xRx300M

M =3xRx30030x30xR

= 1g = 0.001 kg