NEET chemistry

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Question
CBSEENCH11008296

A gaseous mixture was prepared by taking equal moles of CO and N2, If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is

  • 0.8 atm

  • 0.9 atm

  • 1 atm

  • 0.5 atm

Solution

C.

1 atm

because space CO presuperscript straight n space equals space straight N presuperscript straight n subscript 2
therefore space straight p subscript CO space equals space straight p subscript straight N subscript 2 end subscript

Given comma space straight p subscript CO space plus space straight p subscript straight N subscript 2 space end subscript equals space 1 space atm

Or space space 2 straight p subscript straight N subscript 2 end subscript space equals space 1 space atm

straight p subscript straight N subscript 2 end subscript space equals space 0.5 space atm

Sponsor Area

Question
CBSEENCH12011026

If x is the amount of adsorbate and m is the amount of adsorbent, which of the following relations is not related to adsorption process?

  • straight x over straight m space equals space straight f left parenthesis straight T right parenthesis space at space constant space straight p
  • p = f(T) at constant (x/m)

  • straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction
  • straight x over straight m space equals space straight f space left parenthesis straight p right parenthesis space at space constant space left parenthesis straight T right parenthesis

Solution

C.

straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction

The correct relation is  straight x over straight m space proportional to fraction numerator space straight p over denominator straight T end fraction

Question
CBSEENCH11008297

Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals,
NO subscript 2 superscript minus comma space NO subscript 3 superscript minus comma space NH subscript 2 superscript minus comma space NH subscript 4 superscript plus comma space SCN to the power of minus ?

  • NH subscript 4 superscript plus space and space NO subscript 3 superscript minus
  • SCN to the power of minus space and space NH subscript 2 superscript minus
  • NO subscript 2 superscript minus space and space NH subscript 2 superscript minus
  • NO subscript 2 superscript minus space and space NO subscript 3 superscript minus

Solution

D.

NO subscript 2 superscript minus space and space NO subscript 3 superscript minus

Hybridization of the given molecule is 
NO subscript 2 superscript minus space rightwards arrow space sp squared
NO subscript 3 superscript minus space rightwards arrow space sp squared
NH subscript 2 superscript plus space rightwards arrow space sp cubed
NH subscript 4 superscript minus space rightwards arrow space sp cubed
SCN to the power of plus space rightwards arrow sp
therefore, NO subscript 2 superscript minus space and space NO subscript 3 superscript minus both have the same hybridization.

Question
CBSEENCH11008298

If the enthalpy change for the transition of liquid water to steam is 30kJ mol-1 at 27oC, the entropy change for the process would be. 

  • 1.0 J mol- K-1

  • 0.1 J mol-1 K-1

  • 100 J  mol-1 K-1

  • 10 J mol-1 K-1

Solution

C.

100 J  mol-1 K-1

increment straight G to the power of straight o space equals space increment straight H to the power of straight o space minus space straight T increment straight S to the power of straight o

Given comma space increment straight H subscript vap space equals space 30 space kJ space mol to the power of negative 1 end exponent

increment straight G to the power of straight o space equals space 0 space at space equilibrium comma

increment straight S subscript vap space equals space fraction numerator increment straight H subscript vap over denominator straight T end fraction

space equals space fraction numerator 30 space straight x space 10 cubed space straight x space straight J space mol to the power of negative 1 end exponent over denominator 300 space straight K end fraction
space equals space 100 space straight J space mol to the power of negative 1 end exponent space straight K to the power of negative 1 end exponent

Question
CBSEENCH11008299

The Correct order of increasing bond length of C - H, C-O, C - C and C = C is


  • C - C < C=C < C - O < C - H

  • C - O < C - H < C - C < C = C

  • C - H < C - O <  C - C < C= C

  • C - H  < C = C < C - O < C - C

Solution

D.

C - H  < C = C < C - O < C - C

C - H: 0.109 nm
C = C : 0.134 nm
C - O: 0.143 nm
C - C : 0.154 nm
Therefore, Bond length order is 
C - H < C = C < C- O < C - C