NEET chemistry

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Question
CBSEENCH11008337

Which has the maximum number of molecules among the following?

  • 44 g CO2

  • 48 g O3

  • 8 g H2

  • 64 g SO2

Solution

A.

44 g CO2

44 g CO2 = 1 mol CO2 = NA molecules of CO2

48 g O3 = 1 mol O3 = NA molecules of O3
8 g H2 = 4 mol H2 = 4 x NA molecules of H2
64 g SO2 = 1 mol SO2 = NA molecules of SO2

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Question
CBSEENCH11008338

According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

  • n= 6 to n = 1

  • n = 5 to n = 4

  • n= 6 to n = 5 

  • n = 5 to n = 3 

Solution

C.

n= 6 to n = 5 

increment straight E space proportional to open square brackets space fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets space comma space where space straight n subscript 2 greater than straight n subscript 1
Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.

Question
CBSEENCH11008339

Consider the following processes Δ H (kJ/mol)

1/2 A →                               +150
3B   → 2 C + D                     -125
E + A  → 2D                        +350

For  B + D   → E + 2C, ΔH will be 

  • 525 kJ/mol

  • -175 kJ/mol

  • -325 kJ /mol

  • 325 kJ/mol

Solution

B.

-175 kJ/mol

1 half space straight A space space rightwards arrow space straight B semicolon space space space space space space space space space space space space space space space space space space space space space space increment space equals space 150 space space kJ divided by space mol space space.. space left parenthesis straight i right parenthesis

3 straight B space rightwards arrow space 2 space straight C space plus space straight D space semicolon space space space space space space space space space space space increment space straight H space equals space minus space 125 space kJ divided by mol space... space left parenthesis ii right parenthesis

straight E space plus straight A space rightwards arrow space 2 straight D semicolon space space space space space space space space space space space space space space space space increment space straight H space equals space plus 350 space kJ divided by mol space.... space left parenthesis iii right parenthesis
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By space left square bracket space 2 space straight x space left parenthesis straight i right parenthesis space plus space left parenthesis ii right parenthesis space minus space left parenthesis iii right parenthesis comma space we space have


space straight B space plus space straight D space space rightwards arrow space straight E space plus space thin space 2 straight C
therefore space increment straight H space equals space 150 space straight x space 2 space space plus space left parenthesis negative 125 right parenthesis space minus space 350
equals space minus 175 space kJ divided by mol

Question
CBSEENCH11008340

Which of the following structures is the most preferred and hence of lowest energy of SO3

Solution

D.

Formal charges help in the selection of the lowest energy structure from a number of possible energy structure from a number of possible Lewis structures for a given species. Generally, the lowest energy structure is the one with the smallest formal charges on the atoms. 
Formal charge on an atom: = total number of valence electrons- non - bonding electrons - 1/2 x bonding electrons.
For lewis structure of SO3

Formal charge on three O atoms
equals space 6 space minus space 4 space minus 1 half straight x space 4 space equals space 0

Question
CBSEENCH11008342

A bubble of air is underwater at temperature 15o C  and the pressure 1.5 bar. If the bubbles rises to the surface where the temperature is 25o C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

  • Volume will become greater by a factor of 1.6 

  • Volume will become  greater by a factor of 1.1

  • Volume will become  smaller by a factor of 0.70

  • Volume will become greater by a factor of 2.9

Solution

A.

Volume will become greater by a factor of 1.6 

fraction numerator straight p subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight p subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction space space space space left parenthesis ideal space gas space equation right parenthesis
or
fraction numerator 1.5 space straight x space straight V subscript 1 over denominator 288 end fraction space equals space fraction numerator 1 space straight x space straight V subscript 2 over denominator 298 end fraction

therefore space space straight V subscript 2 space equals space 1.55 space straight V subscript 1 space
i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.