Subject

Chemistry

Class

NEET Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

An Organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be

  • CH3O

  • CH2O

  • CHO

  • CHO


A.

CH3O

Element % At. wt Relative no. of atoms Simplest ratio of atoms
C 38.71 12 3.23 3.23/3.23 = 1
H 9.67 1 9.67 9.67/3.23 = 3
O 51.62 16 3.23 3.23/3.23
 
Hence empirical formula is CH3O.
480 Views

12.

Four diatomic solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH?

  • straight O subscript 2 superscript minus space less than thin space NO space less than thin space straight C subscript 2 superscript 2 minus end superscript space less than thin space He subscript 2 superscript plus
  • NO less than thin space straight C subscript 2 superscript 2 minus end superscript space less than thin space straight O subscript 2 superscript minus space less than thin space He subscript 2 superscript plus
  • straight C subscript 2 superscript 2 minus end superscript space less than space He subscript 2 superscript plus less than thin space NO thin space less than thin space straight O subscript 2 superscript minus
  • straight C subscript 2 superscript 2 minus end superscript space less than space He subscript 2 superscript plus less than thin space NO thin space less than thin space straight O subscript 2 superscript minus

D.

straight C subscript 2 superscript 2 minus end superscript space less than space He subscript 2 superscript plus less than thin space NO thin space less than thin space straight O subscript 2 superscript minus
Bond space order space equals space fraction numerator straight N subscript straight b minus straight N subscript straight a over denominator 2 end fraction
He subscript 2 superscript plus colon space left parenthesis straight sigma space 1 right parenthesis squared space left parenthesis straight sigma asterisk times 1 straight s right parenthesis to the power of 1
Bond space order space equals space fraction numerator 2 minus 1 over denominator 2 end fraction space equals space 1 half
straight C subscript 2 superscript 2 minus end superscript colon
KK space left parenthesis straight sigma 2 straight s right parenthesis squared thin space left parenthesis straight sigma asterisk times 2 straight s right parenthesis squared left parenthesis straight pi 2 straight p subscript straight x right parenthesis squared left parenthesis straight pi 2 straight p subscript straight y right parenthesis squared left parenthesis straight sigma 2 straight p subscript straight z right parenthesis squared
Bond space order space space equals fraction numerator 8 minus 5 over denominator 2 end fraction space equals space 1 1 half
NO colon space KK space left parenthesis straight sigma 2 straight s squared right parenthesis left parenthesis straight sigma asterisk times 2 straight s right parenthesis squared left parenthesis straight sigma 2 straight p subscript straight z right parenthesis squared left parenthesis straight pi 2 straight p subscript straight x right parenthesis squared space left parenthesis straight pi 2 straight p subscript straight y right parenthesis squared left parenthesis straight pi asterisk times 2 straight p subscript straight x right parenthesis to the power of 1
Bond space order space equals space fraction numerator 8 minus 3 over denominator 2 end fraction space equals space 2 1 half
Hence comma space the space order space of space increasing space bond space order space is space as space colon
He subscript 2 superscript plus space less than space straight O subscript 2 superscript minus space less than space NO space less than thin space straight C subscript 2 superscript 2 minus end superscript
326 Views

13.

The dissociation equilibrium  of gas AB2 can be represented as

2AB2 (g) ⇌ 2AB (g) + B2 (g)
The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation  (x) with equilibrium constant Kp and total pressure p is 

  • (2Kp / p)

  • (2Kp/p)1/3

  • (2Kp/p)1/2

  • (2Kp/p)1/2


D.

(2Kp/p)1/2

Initial              1                   0               0
at equ        2(1-x)              2x               x

where,  x = degree of dissociation
Total moles at equilibrium = 2-2x + 2x + x
 = ( 2 + x)

straight p subscript AB subscript 2 end subscript space equals space fraction numerator 2 space left parenthesis 1 minus straight x right parenthesis straight p over denominator left parenthesis 2 plus straight x right parenthesis end fraction
straight p subscript AB space equals space fraction numerator 2 xp over denominator left parenthesis space 2 space plus straight x right parenthesis end fraction
straight p subscript straight B subscript 2 space end subscript space equals space fraction numerator xp over denominator space left parenthesis space 2 space plus straight x right parenthesis end fraction
straight K subscript straight p space equals space fraction numerator left parenthesis straight p subscript AB right parenthesis squared left parenthesis straight p subscript straight B subscript 2 right parenthesis over denominator left parenthesis straight p subscript AB subscript 2 end subscript right parenthesis end fraction
space equals space fraction numerator open parentheses begin display style fraction numerator 2 xp over denominator 2 plus straight x end fraction end style close parentheses squared open parentheses begin display style fraction numerator straight x over denominator 2 space plus straight x end fraction end style straight p close parentheses over denominator open parentheses begin display style fraction numerator 2 left parenthesis 1 minus straight x right parenthesis over denominator 2 plus straight x end fraction end style straight p close parentheses end fraction
equals space fraction numerator straight x cubed straight p over denominator left parenthesis 2 plus straight x right parenthesis left parenthesis 1 minus straight x right parenthesis squared end fraction
open square brackets therefore space straight x less than less than less than space 1 space and space 2 space so comma space left parenthesis 1 minus straight x right parenthesis space almost equal to 1 comma space left parenthesis 2 plus straight x right parenthesis space almost equal to space 2 close square brackets
equals fraction numerator straight x cubed straight p over denominator 2 end fraction
straight x space equals open parentheses fraction numerator 2 straight K subscript straight p over denominator straight p end fraction close parentheses to the power of 1 divided by 3 end exponent

1863 Views

14.

Equimolar solutions of the following  were prepared in water separately which one of the solutions will record the highest pH? 

  • SrCl2

  • BaCl2

  • MgCl2

  • MgCl2


B.

BaCl2

As the basic nature increases

pH  of base > 7
pH of acid < 7

In alkaline earth metals on moving downward the size of cation increases, thus basicity increases. Hence, the increasing order or basicity is as:

MgCl2 < CaCl2 < SnCl2< BaCl2
Therefore, the solution of BaCl2 will record the highest pH.

707 Views

15.

The correct order of increasing bond angles in the following triatomic species is 

  • NO subscript 2 superscript minus space less than thin space NO subscript 2 superscript plus space thin space less than space NO subscript 2
  • NO subscript 2 superscript minus space less than thin space NO subscript 2 space less than thin space NO subscript 2 superscript plus
  • NO subscript 2 superscript plus space less than thin space NO subscript 2 space less than thin space NO subscript 2 superscript minus
  • NO subscript 2 superscript plus space less than thin space NO subscript 2 space less than thin space NO subscript 2 superscript minus

B.

NO subscript 2 superscript minus space less than thin space NO subscript 2 space less than thin space NO subscript 2 superscript plus

As the number of lone pair of electrons increases, bond angle decreases. NO2+ ion is isoelectronic with the CO2 molecule. It is a linear ion and its central atom (N+) undergoes sp-hybridisation, hence bond angle is 180o.

In NO2- ion, N -atom undergoes sp2 hybridisation. The angle between hybrid orbital should be 120o but one lone pair of electrons is lying on N- atom, hence bond angle decreases to 115o.

In NO2 molecule, N -atom has one unpaired electron in a sp2-hybrid orbital. The bond angle should be 120o but actually, it is 132o. It may be due to one unpaired electron in a sp2-hybrid orbital.

Therefore, the increasing order of bond angles.

stack NO subscript 2 superscript minus space with 115 to the power of 0 below less than space stack NO subscript 2 with 132 to the power of 0 below space less than space stack NO subscript 2 superscript plus with 180 to the power of 0 below


782 Views

16.

What volume of oxygen gas (O2) measured at 0oC and 1 atm, is needed to burn completely 1 L of propane gas (C3H8)  measured under the same conditions?

  • 7 L

  • 6 L 

  • 5 L 

  • 5 L 


C.

5 L 

Volume of a gas at STP = 22.4 L 

C38 +    5O2 → 3CO2 + 4H2O
22.4 L       5 x 22.4 L

Therefore,
to burn 22.4 L C3H8 the oxygen required is 
 = 5 x 22.4 L 

To burn 1 L C3H8  the oxygen required will be  = 5 x 22.4 / 22.4  = 5 L 

458 Views

17.

Base strength of 

1. space straight H subscript 3 straight C straight C with minus on top straight H subscript 2
2. space straight H subscript 2 straight C equals straight C with minus on top straight H
3. space straight H minus space straight C identical to straight C with minus on top

is in the order of 

  • (2) > (1) > (3)

  • (3) > (2) > (1) 

  • (1) > (3)> (2) 

  • (1) > (3)> (2) 


D.

(1) > (3)> (2) 

Stronger the acid, weaker is its conjugate base. 
The strength of their conjugate acids are in the order: 

HC space identical to CH space greater than thin space straight H subscript 2 straight C space equals CH subscript 2 space greater than thin space CH subscript 3 minus CH subscript 3
Therefore, the correct order of strength of their conjugate base is: 

CH subscript 3 CH subscript 2 to the power of minus greater than thin space straight H subscript 2 straight C space equals straight C to the power of minus straight H space greater than thin space HC identical to straight C to the power of minus

1077 Views

18.

Bond dissociation enthalpy of H2, Cl2 and HCl and 434, 242 and 431 kJ mol-1 respectively.  Enthalpy of formation of HCl is 

  • 93 kJ mol-1

  • -245 kJ mol-1

  • -39 kJ mol-1

  • -39 kJ mol-1


C.

-39 kJ mol-1

increment straight H subscript reaction space equals space begin inline style stack sum space Bond space space energy space of space reactant space with space below and space on top end style begin inline style stack sum Bond space space energy space of space product space with space below and space on top end style
Here comma space increment straight H subscript straight H minus straight H end subscript space equals space 434 space kJ space mol to the power of negative 1 end exponent
increment straight H subscript Cl minus space Cl end subscript space equals space 242 space kJ space mol to the power of minus
increment straight H subscript straight H minus Cl end subscript space equals space 431 space kJ space mol to the power of negative 1 end exponent
because space 1 half space straight H subscript 2 space plus 1 half Cl subscript 2 space rightwards arrow HCl
increment straight H subscript reaction space end subscript space equals space 1 half increment straight H subscript straight H minus straight H end subscript space plus space 1 half increment straight H subscript Cl space minus Cl end subscript space minus space increment straight H subscript straight H minus Cl end subscript
equals space 1 half space straight x space 434 space straight x space 1 half space straight x space 242 minus 431
equals space 217 space plus 121 minus 431
equals negative 93 space kJ space mol to the power of negative 1 end exponent
556 Views

19.

How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?

  • 0.044

  • 0.333

  • 0.011

  • 0.011


D.

0.011

The reagent which is present in smaller quantity is called the limiting reagent and the moles of product depends on it and number of moles
 
 =fraction numerator weight over denominator molecular space weight end fraction

PbO with 207.2 plus 16 below space space space space space space space plus stack space 2 space HCl space with 2 space left parenthesis 35.5 space plus 1 right parenthesis below space rightwards arrow space stack PbCl subscript 2 with 207.2 space plus 71 below space space space plus space space straight H subscript 2 straight O
space equals space 223.2 space space space space space space space space space space equals 73 space space space space space space space space space space space space space space space equals 278.2

Here, 1 mole of PbO reacts with 2 moles of HCl, thus PbO is the limiting reagent

because space 223.2 space straight g space PbO space gives space PbCl subscript 2 space equals space 278.2 space straight g
therefore space 6.5 space straight g space PbO space will space give space PbCl subscript 2
space equals space fraction numerator 278.2 over denominator 232.2 end fraction space straight x space 6.5 space straight g
equals space fraction numerator 278.2 space straight x space 6.5 over denominator 223.2 space straight x space 278.2 end fraction space mol

0.029 mol

449 Views

20.

The measurment of the electron position is associated with an uncerrtanity in momentum, which equal to 1 x 10-18 g cm s-1.The uncertanity in electron velocity is,

(mass of an electrons is 9 x 10-28 g) 

  • 1 x 108

  • 1  x 106 cm s-1

  •  1 x 105 cm s-1

  •  1 x 105 cm s-1


A.

1 x 108

482 Views