Thursday, August 19, 2021

P Block Elements: Chemistry Entrance

 


Q.1. Which of the following is a tetrabasic acid ?

A) Orthophosphoric acid

B) Hypo phosphorous acid

C) Metaphosphoric acid

D) Pyrophosphoric acid

Explanation-

Pyrophosphoric acid : straight H subscript 4 straight P subscript 2 straight O subscript 7

(Tetrabasic) table row blank straight O blank blank blank blank row blank cell vertical line vertical line end cell blank blank blank blank row HO minus straight P minus straight O minus row blank vertical line blank blank blank blank row blank OH blank blank blank blank end table table row straight O row cell table row cell vertical line vertical line end cell row straight P row vertical line end table end cell row OH end table table row cell table row blank row blank row minus end table end cell row blank row blank end table table row cell table row blank row blank row OH end table end cell row blank row blank end table 


Q.2. When ammonia is passed over heated copper oxide, the metallic copper is obtained, The reaction shows that ammonia is

A) A dehydrating agent

B) An oxidizing agent

C) A reducing agent

D) A nitrating agent

Explanation-

2 NH with negative 3 on top subscript 3 space plus space 3 stack CuO space with plus 2 on top rightwards arrow with triangle on top straight N with 0 on top subscript 2 space plus space 3 Cu with 0 on top space plus space 3 straight H subscript 2 straight O 


Q.3.  Which is the most amphoteric oxide ?

A) Nablank subscript 2O

B) MgO

C) Alblank subscript 2Oblank subscript 3

D) CaO

Explanation-

Oxide of aluminum is amphoteric in nature, whereas NaO, MgO and CaO are basic in nature.


Q.4. Which of the following statements is incorrect  ?

A) BFblank subscript 3  is the weakest Lewis acid 

B) Ammonal is a mixture of aluminium compounds and is used in bombs 

C) BFblank subscript 3 exists as dimer 

D) BCIblank subscript 3  does not conduct electricity in its molten state 

Explanation-

 BFblank subscript 3 exists as dimer.


Q.5.  Orthoboric acid contains 

A) Pyramidal B O subscript 3 to the power of 3 minus end exponent  units

B) Linear  B O subscript 3 to the power of 3 minus end exponent units 

C) T-shaped  B O subscript 3 to the power of 3 minus end exponent units 

D) Triangular  B O subscript 3 to the power of 3 minus end exponent units    

Explanation-

Orthoboric acid open parentheses H subscript 3 B O subscript 3 close parentheses contains triangular BO subscript 3 superscript 3 minus end superscript units joined together through H-bonding forming a trigonal planar layer structure.


Q.6. Aluminum deposited as vapors on glass forms a good mirror, essentially because 

A) It has better shine than Ag

B) It does not scratch

C) Coating is  much smoother

D) It does not tarnish in air

Explanation-

Aluminum being passive metal does not tarnish in air.


Q.7. In the electrolytic extraction of aluminum fused cryolite is added

A) To make a conduction solution of bauxite

B) To act as reducing agent

C) To increase Al production

D) To protect the anodes

Explanation-

Addition of cryolite to fused alumina during electrolytic extraction make the solution more conducting and low melting.


Q.8.  Which of the metal chlorides has maximum covalent character ?

A) NaCl

B) AlClblank subscript 3

C) CsCl

D) BaClblank subscript 2 

Explanation-

Al to the power of 3 plus end exponent has minimum size and maximum charges, hence forms compound with maximum covalent character among given choice. The covalent character of chlorides increases along a period and decreases down a group according of Fajan's rule.


Q.9. When concentrated H subscript 2 S O subscript 4 is heated with K N O subscript 3, brown fumes evolve. These fumes are due to

A) S O subscript 2

B) S O subscript 3

C) N O subscript 2

D) N O 

Explanation-

KNO subscript 3 space plus space straight H subscript 2 SO subscript 4 rightwards arrow with triangle on top KHSO subscript 4 space plus space HNO subscript 3
4 HNO subscript 3 space rightwards arrow with triangle on top space 4 NO subscript 2 space plus space 2 straight H subscript 2 straight O space plus space straight O subscript 2
space space space space space space space space space space space space space space space space space space space space space Brown 


Q.10. Which one of the following elements is most metallic ?

A) P

B) As

C) Sb

D) Bi

Explanation-

Metallic character increases down the group.

Kinetic Theory Of Gases : Physics Entrance Exam

 



Q.1.  If k is the Boltzmann constant , the average kinetic energy of a gas molecule at absolute temperature T is 

A) k T/2

B) 3 k T/4

C) k T

D) 3 k T/2


Q.2. At room temperature (27degreeC ) the rms speed of the molecules of a certain diatomic gas is found to be 1920 msblank to the power of negative 1 end exponent.The  gas is 

A) Hblank subscript 2

B) Fblank subscript 2

C) Oblank subscript 2

D) CIblank subscript 2 

Explanation-

We know that Ï‘ subscript r m s end subscript space equals space square root of fraction numerator 3 R T over denominator M end fraction end root which gives 

M equals fraction numerator 3 space R T over denominator Ï‘ squared subscript r m s end subscript end fraction equals fraction numerator 3 cross times 8.3 cross times 300 over denominator open parentheses 1920 close parentheses squared end fraction
space space space equals space 2 cross times 10 to the power of negative 3 end exponent space k g space equals space 2 space g

 Since  M = 2 , the gas is hydrogen .


Q.3.  If a g as has f degrees of freedom, the ratio C subscript p divided by C subscript v of the gas is 

A) fraction numerator 1 plus f over denominator 2 end fraction

B) 1 plus f over 2

C) 1 half plus f

D) 1 plus 2 over f 

Explanation-

We know that C subscript Ï‘ = no . of degress of freedom x molar gas constant / 2 or C subscript Ï‘ equals f R divided by 2. space A l s o space C subscript p minus C subscript Ï‘ space equals space R
T h e r e f o r e
C subscript p over C subscript Ï‘ space R over C subscript Ï‘ plus 1 equals fraction numerator 2 R over denominator f space R end fraction plus 1 equals 2 over f plus 1


Q.4.  Two ideal gases at absolute temperature Tblank subscript 1 and Tblank subscript 2 are mixed . There is no loss of energy in this process . If nblank subscript 1 and nblank subscript 2 are the respective number of molecules of the gases , the temperature of the mixture will be 

Afraction numerator n subscript 1 T subscript 1 space plus space n subscript 2 T subscript 2 over denominator n subscript 1 plus n subscript 2 end fraction

B) fraction numerator n subscript 2 T subscript 1 plus n subscript 1 T subscript 2 over denominator n subscript 1 plus n subscript 2 end fraction

C) T subscript 1 plus n subscript 2 over n subscript 1 T subscript 2

D) T subscript 2 plus n subscript 1 over n subscript 2 T subscript 1 

Explanation-

fraction numerator n subscript 1 T subscript 1 space plus space n subscript 2 T subscript 2 over denominator n subscript 1 plus n subscript 2 end fraction


Q.5.  Four molecules of a gas have speeds 1,2,3 and 4 km s to the power of negative 1 end exponent . The value of the root mean square speed of the gas molecule is 

A)  1 half space square root of 15 space end root space k m space s to the power of negative 1 end exponent

B) 1 half square root of 10 space k m space s to the power of negative 1 end exponent

C) 2.5 space k m space s to the power of negative 1 end exponent

Dsquare root of 15 divided by 2 end root space k m space s to the power of negative 1 end exponent 

Explanation-

The rms speed = open square brackets fraction numerator 1 squared plus 2 squared plus 3 squared plus 4 squared over denominator 4 end fraction close square brackets to the power of 1 divided by 2 end exponent
equals space square root of 15 over 2 end root space k m s to the power of negative 1 end exponent


Q.6.   The average kinetic energy of a molecule of a gas at absolute temperature T is proportional to 

A) 1/T

B) square root of T

C) T

D) Tblank squared 

Explanation-

The average kinetic energy of a molecule of a gas at absolute temperature T is given by straight E space equals space 3 over 2 space kT.


Q.7.  The root mean square speed of the molecules of a gas at absolute temperature T is proportional to 

A) 1/T

B) square root of T

C) T

D) Tblank squared 

Explanation-

The root mean square speed of the molecules of a gas is given by straight Ï‘ subscript rms space end subscript equals space square root of fraction numerator 3 space kT over denominator straight m end fraction end root.


Q.8.  E subscript 0  and E subscript h respectively represent the average kinetic energy of a molecule of oxygen and hydrogen. If the two gases are at the same temperature , which of the following statements is true ?

A)  E subscript 0 > E subscript h 

B)  E subscript 0 = E subscript h 

C)  E subscript 0 < E subscript h 

D)  Nothing can be said about the magnitude of  E subscript 0 and E subscript h as the information given is insufficient  

Explanation-

The average kinetic energy of molecule = 1 half kT which is independent of the molecular mass .


Q.9.  The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K , the root mean square speed of the gas molecules is v , then at 480 K it will be 

A) 4 v

B) 2 v

C) v over 2

D) v over 4 

Explanation-

Ï‘ subscript r m s end subscript equals square root of fraction numerator 3 R T over denominator M end fraction end root space o r space Ï‘ subscript r m s end subscript proportional to square root of T . 


Q.10.  A vessel contains 1 mole of Oblank subscript 2 gas ( molar mass 32 ) at a temperature T . The pressure of the gas is P . An identical vessel containing one mole of He gas ( molar mass 4 ) at a temperature  2 T has a pressure of 

A) p over 8

B) P

C) 2 P

D) 8 P

Explanation-

For a gas , PV = nRT . Hence 

                   ( P )open parentheses straight P close parentheses o subscript 2 equals fraction numerator open parentheses 1 space m o l e close parentheses space R T over denominator V end fraction
a n d space left parenthesis P right parenthesis subscript H e end subscript equals fraction numerator open parentheses 1 space m o l e close parentheses R open parentheses 2 T close parentheses over denominator V end fraction
therefore fraction numerator open parentheses P close parentheses subscript H e end subscript over denominator open parentheses P close parentheses o subscript 2 end fraction space equals 2 space o r space left parenthesis P right parenthesis subscript H e end subscript equals 2 open parentheses P close parentheses o subscript 2