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27351.

The numberof carbon atoms presentin human blood ?

Answer»

6
4
3
5

Answer :B
Discussion
27352.

The number of carbon compounds is very large because it :

Answer»

Is tetravalent
Forms DOUBLE and TRIPLE bonds
Is non-metal
Shows catenation

Answer :D
Discussion
27353.

The number of carbon atoms present in a signature , if a signaturewritten by carbon pencil, weighing 1.2 xx 10^(-3) g is:

Answer»

`12.04xx10^(20)`
`6.02xx10^(19)`
`3.01xx 10^(19)`
`6.02xx 10^(20)`

ANSWER :B
Discussion
27354.

The number of carbon atoms per unit eel I of diamond unit cell is:

Answer»

8
6
1
4

Solution :Diamond is like ZNS. In diamond cubic unit cell, there are eight CORNER ATOMS, six face centered atoms and four more atoms INSIDE the strucn1 re. Number of atoms present in a diamond cubic cell
`8XX(1)/(8)xx6xx(1)/(2)xx4xx8`
Discussion
27355.

The number of carbon atoms per unit cell of diamond unit cell is .........

Answer»

6 
1 
4 
8 

SOLUTION :In diamond, half of the total tetrahedral voids are occupied by carbon i.e. 4
Total number of CARBONS per unit CELL
` = (8 xx 1/8) + 4 + (6 xx 1/2) = 8`
Discussion
27356.

The number of carbon atoms per unit cell of diamond is …………………. .

Answer»

8
6
1
4

Solution :Hint : In diamond CARBON forming fcc. Carbon occupies corners and FACE centres and also OCCUPYING half of the tetrahedral voids.
`(N_(c))/(8)+(N_(f))/(2)+4C" atoms in Td voids"`
`((8)/(8))+((6)/(2))+4=8`
Discussion
27357.

The number of carbon atoms per unit cell of diamond unit cell is

Answer»

8
6
1
4

Answer :A
Discussion
27358.

The number of carbon atoms per unit cell of diamond is

Answer»

8
6
1
4

Solution :In DIAMOND CARBON FORMING fcc. Carbon occupies corners and face CENTRES and also occupying half of the tetrahedral voids.
`(N_(c))/(8) + (N_(f))/(2) + 4` C atoms in Td voids
`((8)/(8)) + ((6)/(2)) + 4 = 8`
Discussion
27359.

The number of C - C bonds in Hexamethylena tetramine (CH_(2))_(6)N_(4) or urotropine are

Answer»

9
6
4
0

Answer :D
Discussion
27360.

The number of carbon atoms in a chain can be increased with ______ reaction.

Answer»

Grignard
Cannizaro reaction
HVZ
Clemmensen

Answer :A
Discussion
27361.

The number of bridging CO groups in [Co_2(CO)_8] and [Fe(CO)_5] are respectively

Answer»

2, 0
1, 1
1, 0
2, 1

Answer :A
Discussion
27362.

The number of bridging carbonyls respectively in Mn_2(CO)_10 and CO_2(CO)_8 are

Answer»

0,0
1,1
1,0
0,2

Answer :D
Discussion
27363.

The number of Bravis lattices possible for the crystal system monoclinic is

Answer»

2
1
4
3

Answer :A
Discussion
27364.

The number of bonds between sulphur and oxygen atoms in S_2Og_2 and number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are :

Answer»

8 and 6
4 and 6
8 and 8
4 and 8

Solution :`-O=underset(O)underset(||)OVERSET(O)overset(||)S - O - O - underset(O)underset(|)overset(O)overset(||)S-O-`
`S_(2)O_(8)^(2-)`
Discussion
27365.

The number of bond pair and lone pair on oxygen atom in ether are respectively

Answer»

1 and 2
2 and 1
2 and 2
1 and 3

Answer :C
Discussion
27366.

The number of Bravais lattices in a cubic crystal is:

Answer»

3
1
4
14

Answer :A
Discussion
27367.

The number of beta-particles emitted in the following nuclear reaction is

Answer»

Both Assertion and REASON are CORRECT and Reason is the correct EXPLANATION of the Assertion
Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion
Assertion is correct but Reason is incorrect
Assertion is incorrect but Reason is correct

Answer :4
Discussion
27368.

The number of beta - particle emitted during the change ""_(a) X^(c) to ""_(a) Y^(b) is :

Answer»

`(a- b)/(4)`
`d + [ (a- b)/(2)]`
`d- a + [(C- d)/(2)]`
`d + [ (a- b)/(2)] - c`

SOLUTION :`""_(a) X^(c) to ""_(d) Y^(b) + x (""_(2) He^(4)) + y (""_(-1) beta^(0)) , x = ((c- b)/(4))`
`a = d + 2x - y , y = d + 2x -a , y = d + 2 ((c - b)/(4)) - a , , y =[d - a + (c- b)/(2)]`
Discussion
27369.

The number of basic types of unit cells among the crystals is :

Answer»

Eight
Fourteen
SEVEN
Ten

Solution :There are seven BASIC UNIT CELLS.
Discussion
27370.

The number of basic crystal systems is

Answer»

7
8
6
4

Answer :A
Discussion
27371.

The number of atoms which are sp^(2) hybridised in aniline is

Answer»


SOLUTION :No of ATOMS `SP^(2)`
Discussion
27372.

The number of basic crystal systems are:

Answer»

7
8
6
4

Answer :A
Discussion
27373.

The number of atoms present in one molecule of sulphur is:

Answer»

2
4
6
8

Answer :D
Discussion
27374.

The number of atoms present in one molecule of a substance is called its…………… .

Answer»

SOLUTION :ATOMICITY
Discussion
27375.

The number of atoms present in a simple cubic unit cell are-

Answer»

4
3
2
1

Answer :D
Discussion
27376.

The number of atoms present in one mole of an element is equal to Avogadro's number. Which of he following element contains the greatest number of atoms?

Answer»

4g He
46g Na
0.40 g Ca
12g He

Solution :`"4 g He"=(46)/(23)"mole = 1 mole"=N_(A)" atoms"`
`"46 g Na"=(46)/(23)"mole = 0.5 moles"=(N_(A))/(2)"atoms"`
`"0.40 g Ca"=(0.40)/(40)"mole"=10^(-2)" mole"=10^(-2)N_(A)" pr "(N_(A))/(100)" atoms"`
`"12 g He"=(12)/(4)" mole =3 mole "=3N_(A)" atoms"`
THUS, 12 g He contains MAXIMUM NUMBER of atoms.
Discussion
27377.

The number of atoms present in a simple cubic unit cell are:

Answer»

4
3
2
1

Answer :D
Discussion
27378.

The number of atoms present in a hexagonal close packed unit cell is

Answer»

4
6
8
12

Answer :B
Discussion
27379.

The number of atoms per unit cell in a solid is 4. What type of cubic unit cell is it ?

Answer»

SOLUTION :Face-centered
Discussion
27380.

The number of atoms per unit cell in a simple cubic arrangement is :

Answer»

1
8
4
2

Answer :A
Discussion
27381.

The number of atoms per unit cell in a face-centred cube is:

Answer»

2
3
4
14

Answer :C
Discussion
27382.

The number of atoms per unit cell in a simple cube, face-centred cube and body-centred cube are respectively:

Answer»

1,4,2
1,2,4
8,14,9
8,4,2

Answer :A
Discussion
27383.

The number of atoms per unit cell in a simple cube, face centred cube and body centred cube are rrepectively:

Answer»

2
3
4
14

Answer :C
Discussion
27384.

The number of atoms per unit cell in a body centred cubic arrangement is :

Answer»

1
3
4
6

Answer :B
Discussion
27385.

The number of atoms of oxygen present in 11.2 L of ozone at N.T.P. are :

Answer»

`3.01 XX 10^23`
`6.02 xx 10^23`
`9.03 xx 10^23`
`1.20 xx 10^24`

Solution :22.4 L of OZONE at N.T.P.` = 6.02 xx 10^23`
molecules of `O_3 = 3 xx 6.02 xx 10^23` ATOMS of O 11.2 L of ozone at N.T.P.
` = (3 xx 6.02 xx 10^23)/(22.4) xx 11.2 = 9.03 xx 10^23`
Discussion
27386.

The number of atoms (n) contained within a cubic cell is :

Answer»

1
2
3
4

Answer :B
Discussion
27387.

The number of atoms/molecules present in one body centred cubic unit cell is:

Answer»

1
2
4
6

Answer :B
Discussion
27388.

The number of atoms/molecules contained in one face centred cubic unit cell of a monoatomic substance is-

Answer»

4
6
8
12

Answer :A
Discussion
27389.

The number of atoms/molecules contained in one face centred cubic unit cell of a monoatomic substance is:

Answer»

4
6
8
12

Answer :A
Discussion
27390.

The number of oxygen atoms bonded to one phosphorous atom in P_(4)O_(10) is

Answer»


Solution :The STARED ATOMS in the above STRUCTURE forms the larget possible hetero cyclic ring which CONTAINS 2 - atoms (4 - P.s + 4 - O. s)
Discussion
27391.

The number of atoms in the cyclic structure of D-fructose is ____________

Answer»

5
6
4
7

Solution :5 ATOMS in the RING
Discussion
27392.

The number of atoms in fcc unit cell is ………………………………

Answer»

2
4
6
8

Answer :B
Discussion
27393.

The number of atoms in HCP unit cell is ________

Answer»

4
8
12
6

Answer :D
Discussion
27394.

The number of atoms in 52 u of He is:

Answer»

`3.1 xx 10^25`
`7.8 xx 10^23`
`13`
`103`

SOLUTION :4 amu of He = 1 ATOM
53 amu of He `= 1/4 xx 52 = 13 ` atoms
Discussion
27395.

The number of atoms in 2.4 g of body centred cubic crystal with length 200 pm is (density=10 g cm^(-3) ,NA =6xx10^(22) atoms/mol)

Answer»

`6 XX 10^(22)`
`6 xx 10^(20)`
`6 xx 10^(23)`
`6 xx 10^(19)`

Solution :(a): Density, `RHO` = 10 g `cm^(-3)`,
`N_(A) = 6 xx 10^(23)` ATOMS/mol, a= 200 pm, For bcc, Z=2
`rho = (ZxxM)/(a^(3)xxN_(a))`
10 g `cm^(-3) = (2xxM)/((200xx10^(10)cm)^(3)xx(6xx10^(23)mol^(-1)))`
M = 24 g `mol^(-1)`
24 g of element contains = `6xx10^(23)` atoms
`:.` 2.4 g of element contains = `(6xx10^(23))/(24) xx 2.4` atoms
`= 6xx10^(22)` atoms
Discussion
27396.

The number of atoms in 0.1 mole of a triatomic gas is (N_(A)=6.02xx10^(23)"mol"^(-1))

Answer»

`1.800xx10^(22)`
`6.026xx10^(22)`
`1.806xx10^(23)`
`3.600xx10^(23)`

ANSWER :A
Discussion
27397.

the number of atoms in 0.1 mole of a triatomic gas is (N_(A)=6.02xx10^(23) mol^(-1))

Answer»

`1.800xx10^(22)`
`6.026xx10^(22)`
`1.806xx10^(23)`
`3.600xx10^(23)`

SOLUTION :No. of ATOMS = No. of MOLECULES `xx` atomicity
`0.1 N_(A)xx3=0.1xx6.02xx10^(23)xx3=1.806xx10^(23)`
Discussion
27398.

The number of atoms in 0.1 mol triatomic gas is: (N_(A)=6.02xx10^(23)"mol"^(-1))

Answer»

`6.026xx10^(22)`
`1.806xx10^(23)`
`3.6xx10^(23)`
`1.8xx10^(22)`

SOLUTION :No. of ATOMS `=0.1xx3xx6.02xx10^(23)`
`=1.806xx10^(23)`
Discussion
27399.

The number of atoms in 0.1 mol of a triatomicgas is : (N_A = 6.02 xx 10^23 "mol"^(-1) )

Answer»

`3.600 XX 10^23`
`1.800 xx 10^22`
`6.02 xx 10^22`
`1.806 xx 10^23`

Solution :No. of MOLECULESIN 0.1 MOL` = 0.1 xx 6.02 xx 10^23`
No. of atoms ` = 0.1 xx 6.02 xx 10^23 xx 3`
` = 1.806 xx 10^23`
Discussion
27400.

The number of atoms in 0.004 g of magnesium is closed to :

Answer»

24
`2 XX 10^20`
`10^20`
`6.02 xx 10^23`

ANSWER :C
Discussion