MCQ Questions for Class 12 Engineering Chemistry The Solid State Quiz 5

The total number of lattice arrangements in different crystal systems is:

0%
$14$
0%
$8$
0%
$3$
0%
$7$

Which of the following is an interstitial compound?

0%
$PdCl_2$
0%
$NH_3$
0%
$ZnO$
0%
$TiH$

Explanation

An interstitial compound is a compound that is formed when an atom with a small enough radius sits in an interstitial hole in a metal lattice. Examples of small atoms are boron, carbon and nitrogen. These compounds play an important role in industries. Examples:

$TiH$ ,

$Fe_3H$ and

$Mn_4N$

Which primitive unit cell has unequal edge lengths (

$a \neq b \neq c$ ) and all axial angles different from 90?

0%
Tetragonal
0%
Hexagonal
0%
Monoclinic
0%
Triclinic

Explanation

In Triclinic cell

$a \neq b \neq c\ \&\ \alpha \neq \beta \neq \gamma \neq 90^o$ .

What is the number of ions per

$cc$ in the crystal of

$NaCl$ at normal temperature?

0%
$10^{6}\ ions$
0%
$10^{22}\ ions$
0%
$10^{16}\ ions$
0%
$10^{8}\ ions$

Explanation

Number of ions per cc in crystal of

$NaCl$ at normal temperature is

${10}^{22}$ ions

Which of the following have maximum lattice energy :

0%
$LiF$
0%
$CsCl$
0%
$KBr$
0%
$NaCl$

Explanation

$L.E$

$\alpha$

$|Z^+||Z^-|$

$\alpha$

$\dfrac{1}{Size}$ of ion
So

$LiF$ should have
Highest

$L.E$

In a cube, X atoms are present at corners and Y atoms are at face centres of a cube. The number of X and Y are:

0%
$8,8$
0%
$6,8$
0%
$8,6$
0%
$6,6$

Explanation

No. of corner position in a cube= 8
No. of face centres present in a cube=6

$X=8, Y=6$

$CaF_2$ lattice coordination number of

$Ca^{+2}$ &

$F^-$ is :

0%
$4,4$
0%
$8, 8$
0%
$4, 8$
0%
$8, 4$

Explanation

$Ca^{+2}$ occupy at

$FCC$ lattice site and

$F-$ occupy at tetrahedral void so

$C.N.$ of

$Ca^{+2} = 8$ and

$F- C.N. = 4$

The type of defect in

$NaCl$ crystal will be -

0%
point defect
0%
interstitial defect
0%
vacancy defect
0%
impurity defect

An element X (A.w =80 g/mol) having fcc structure.Calculate the number of unit cells in 8g of X:

0%
$0.4 \times N_4$
0%
$0.1 \times N_4$
0%
$4 \times N_A$
0%
$N_A /40$

A compound is formed by cation

$C$ and anion

$A$ . The anions form hexagonal close packed (hcp) lattice and the cations occupy

$75\%$ of octahedral voids. The formula of the compound is :

0%
$C_2A_3$
0%
$C_3A_2$
0%
$C_3A_4$
0%
$C_4A_3$

Explanation

Anions (A) are in hcp, so the number of anions

$(A)=6$

Cations (C) are in

$75\% \,O.V.,$ so number of cations (C)

$=6\times \dfrac{75}{100}$

$=\dfrac{18}{4}$

$=\dfrac{9}{2}$

$\Rightarrow$ So, the formula of compound will be

$C_{\frac{9}{2}}A_6\Rightarrow C_9A_{12}$

$C_9A_{12}\Rightarrow C_3A_4$

In tetragonal crystal system, which of the following is true?

0%
All axial lengths and all axial angles are equal
0%
All three axial lengths are equal
0%
All three axial angles are equal
0%
Two axial angles are equal but the third is different

Explanation

In tetragonal system

$a=b\ne c,$ and

$\alpha =\beta =\gamma ={90}^{o}$

Here all three axial angles are equal but all lengths in tetragonal system are not same.

Hence, option

$C$ is correct.

Which one of the following is correct?

0%
Schottky defect lowers the density.
0%
Frenkel defect increases the dielectric constant of the crystals.
0%
Stoichiometric defects make the crystals good electrical conductors.
0%
All of the above.

Which has a minimum percent volume occupied by the spheres?

0%
Simple cubic
0%
Body-centred cubic
0%
Hexagonal closed packed
0%
Cubic closed packed

Explanation

The edge length or side of the cube,

$a$ and the radius of each particle,

$r$ are related as

$a = 2r$ .
The volume of the cubic unit cell

$= a^{3} = (2r)^{3} = 8r^{3}$
A simple cubic unit cell contains only

$1$ atom.
The volume of the occupied space

$=\dfrac { 4\pi { r }^{ 3 } }{ 3 }$

$\therefore \text{Packing efficiency }$

$=\dfrac{\text{Volume of one atom}}{\text{Volume of cubic unit cell}}\times 100 =52.4\%$

Which of the following does not represent a type of crystal system?

0%
Triclinic
0%
Monoclinic
0%
Rhombohedral
0%
Isotropical

An element (atomic mass

$=100$ amu) having a bcc structure has a unit cell edge equal to

$400$ pm. The density (in g cm

$^{-3}$ ) of the element is:

0%
$10.376$
0%
$5.200$
0%
$7.289$
0%
$2.144$

Explanation

The formula for calculating density is given below:

$d=\dfrac { z\times M }{ { N }_{ A }\times { a }^{ 3 } }$

where,

$d=$ density of the unit cell

$z=$ effective number of atoms in the bcc unit cell

$=2$

$M=$ molar mass

$=100\ g/mol$

$N_A=$ Avogadro number

$a^3=$ volume of the unit cell

$=(400\times { 10 }^{ -10 })^3cm ^{ 3 }$

Substituting the value, we get

$d=\dfrac { 2\times 100 }{ { 6\times 10 }^{ 23 }\times { (400\times { 10 }^{ -10 }cm) }^{ 3 } } =\ 5.2\ g .cm^{-3}$

The number of basic crystal system present in the solid structure is:

0%
$7$
0%
$8$
0%
$6$
0%
$4$

Amorphous solids are:

0%
Solid substances in real sense
0%
Liquids in real sense
0%
Supercooled liquids
0%
Substances with definite melting point

$TiO_{2}$ is a well known example of :

0%
triclinic system
0%
tetragonal system
0%
monoclinic System
0%
none of the above

Explanation

Tetragonal crystal lattice result from stretching a cubic lattice along with one of its lattice vectors so that the cube becomes a rectangular prism with a square base and a height.

The titanium cations have a coordination number of

$6$ meaning they are surrounded by an octahedron of

$6$ oxygen atoms. The oxygen anions have a co-ordination number of

$3$ resulting in trigonal planar coordination. Rutile also shows a screw axis when its octahedral are viewed sequentially.

A metal crystallises as body centred cubic lattice with the edge length of unit cell equal to

$0.304$ nm. If the molar mass of the metal is

$50.3$ g mol

$^{-1}$ , its density (in g cm

$^{-3}$ ) is :

0%
$5.945$
0%
$2.9725$
0%
$8.9175$
0%
$4.458$

Explanation

The expression for density is given below:

Z=2 for BCC, M= 50.3 g/ mol,

$a= 0.304\ nm = 0.304 \times 10^{-7} cm$ and

$N_A= 6.0 \times 10^{23}$

$d=\dfrac { zM }{ { N }_{ A }{ a }^{ 3 } }$

$=\dfrac { 2\times 50.3 }{ { 6\times 10 }^{ 23 }\times { (0.304\times { 10 }^{ -7 }) }^{ 3 } } =\quad 5.945$ g cm

$^{-3}$

Option A is correct.

Percentage of free space in cubic close-packed structure and in body centred packed structure are, respectively:

0%
$30\%$ and $26\%$
0%
$26\%$ and $32\%$
0%
$32\%$ and $48\%$
0%
$48\%$ and $26\%$

Hydrogen occupies the _________ holes, while carbon and nitrogen occupy __________ holes.

0%
tetrahedral and octahedral
0%
octahedral and tetrahedral
0%
octahedral and octahedral
0%
tetrahedral and tetrahedral

Explanation

The size of the hole in tetrahedral voids is smaller than the size of the hole in octahedral voids. The size of the H atom is smaller than the size of C and N atoms. Hence, hydrogen occupies tetrahedral voids and carbon and nitrogen occupy octahedral voids.

Hence, the correct option is

$A$

Aluminium has fcc structure. The edge length of the unit cell is

$404$ pm. If the density of the metal is

$2.7$ g cm

$^{-3}$ , the molar mass (in gmol

$^{-1}$ ) of Al atom is :

0%
$28.20$
0%
$30.40$
0%
$26.80$
0%
$25.20$

Explanation

The expression for density is as follows:

$d=\dfrac { ZM }{ { N }_{ A }{ a }^{ 3 } }$

where

$Z = 4$ for fcc

$d=\dfrac { 4\times M}{ { 6.023\times 10 }^{ 23 }\times { (404\times { 10 }^{ -10 }) }^{ 3 } } = 2.7$ g cm

$^{-3}$

$\implies M=26.80$ g mol

$^{-1}$

Hence , the molecular mass is

$26.80$ gmol

$^{-1}$ .

An element crystallises in a bcc lattice. Nearest neighbours and percentage of volume occupied by spheres in the unit cell are :

0%
$8,\ 74\%$
0%
$8,\ 68\%$
0%
$6,\ 74\%$
0%
$6,\ 68\%$

Explanation

There are

$8$ nearest neighbours in BCC. Percentage of volume occupied by spheres in the unit cell of BCC is

$68\%$ .

Body diagonal of a cube is

$866$ pm. Its edge length would be:

0%
$408$ pm
0%
$1000$ pm
0%
$500$ pm
0%
$600$ pm

Explanation

Body diagonal =

$\sqrt 3 a$

$\sqrt { 3 } a=\quad 866\quad pm\\ a\quad =\quad 500\quad pm\\$

The effective radius of an iron atom is

$1.42\ \mathring A$ . It has FCC structure. Calculate its density (in g.cm

$^{-3}$ ) if it is given that the atomic mass of

$Fe$ is

$56$ amu.

0%
$2.47$
0%
$5.74$
0%
$8.24$
0%
None of the above

Explanation

$d = \dfrac{ zM}{N_A \times V }$
For FCC ,

$z=4$

$a = 2\sqrt 2 r$

$\implies a = 4.02 \times 10^{-8}$ cm

$d = \dfrac{ 4\times 56}{6.023 \times 10^{23} \times ( 4.02 \times 10^{-8})^3 }$

$d = \dfrac{ 4\times 56}{6.023 \times 64.96 \times 10^{-1} }$

$d = 5.74$

$g cm^{-3}$

A metal crystallizes in two cubic phases, fcc and bcc whose unit cell lengths are

$3.5 \mathring A$ and

$3.0 \mathring A$ respectively. The ratio of density of fcc and bcc is :

0%
$2$
0%
$1.26$
0%
$3.34$
0%
$1.8$

Explanation

The formula for the calculation of density is given below:

$\text{Density}= \displaystyle\frac { Z \times M} {N_0 \times a^3 \times 10^{-30}}$ g cm

$^{-3}$

$Z$ is the number of formula units in the unit cell.

$M$ is the formula mass of the substance.

$a$ is the length (in pm) of the unit cell.

$N_0$ is Avogadro's number.

For FCC,

$Z= 4, \ a=3.5 \mathring A=350$ pm

$D_{fcc}= \displaystyle\frac { 4 \times M} {N_0 \times (350)^3 \times 10^{-30}}$ g cm

$^{-3}$

$......(1)$

For BCC,

$Z= 2,\ a=3.0 \mathring A=300$ pm

$D_{bcc}= \displaystyle\frac { 2 \times M} {N_0 \times (300)^3 \times 10^{-30}}$ g cm

$^{-3}$

$......(2)$

Dividing equation

$(1)$ by equation

$(2)$ , we get

$\displaystyle\frac {D_{fcc}} {{D_{bcc}}}=\displaystyle\frac {\displaystyle\frac { 4 \times M} {N_0 \times (350)^3 \times 10^{-30}}} {\displaystyle\frac { 2 \times M} {N_0 \times (300)^3 \times 10^{-30}}}=\displaystyle\frac {\displaystyle\frac {4}{350^3}}{\displaystyle\frac {2}{300^3}}=1.26$

Edge length of a cube is

$400$ pm. Its body diagonal (in pm) would be:

0%
$566$
0%
$600$
0%
$500$
0%
$693$

Explanation

Edge length of a cube is 400pm. Hence, the body diagonal

$= \sqrt{3}a = \sqrt{3} \times 400 = 693$ pm

An element crystallizes as a face-centred cubic lattice with edge length equal to

$460$ pm. The density (in gcm

$^{-3}$ ) of the element X when molar mass of X atom is

$60$ gmol

$^{-1}$ is:

0%
$4.096$
0%
$2.048$
0%
$6.144$
0%
$3.072$

Explanation

The formula for calculating density is given below:

$d=\dfrac { z\times M }{ { N }_{ A }\times { a }^{ 3 } }$

where,

$d=$ density of the unit cell

$z=$ effective number of atoms in the fcc unit cell

$=4$

$M=$ molar mass

$=60\ g/mol$

$N_A=$ Avogadro number

$a^3=$ volume of the unit cell

$=(460\times { 10 }^{ -10 })^3cm ^{ 3 }$

Substituting the value, we get

$d=\dfrac { z\times M }{ { N }_{ A }\times { a }^{ 3 } }$

$=\dfrac { 4\times 60}{ { 6.023\times 10 }^{ 23 }\times { (460\times { 10 }^{ -10 }) }^{ 3 } } = 4.096$ g cm

$^{-3}$

Percentage of free space in a body-centred cubic unit cell is:

0%
$34%$
0%
$28%$
0%
$30%$
0%
$32%$

Explanation

Packing fraction =

$\dfrac{ \text{Volume occupied by atoms in a unit cell}}{\text {Volume of the unit cell}}$

For Body centered cubic cell , Packing fraction =

$\dfrac{ 2 \times \dfrac{4}{3}\pi \times r^3} {\left({\dfrac {4r}{\sqrt 3}} \right )^2}$ =

$\dfrac {\sqrt3 \pi }{8} = 0.68$

where ,edge length a=

$\dfrac{4r}{\sqrt 3}$

Therefore, volume occupied =

$68\%$

Hence vacant volume =

$32\%$

So correct answer is option

$D$ .

The volume occupied by atoms in a simple cubic unit cell is (edge length

$=$ a) :

0%
$a^{3}$
0%
$\dfrac {4\pi a^{3}}{3}$
0%
$\dfrac {\pi a^{3}}{6}$
0%
$\dfrac {\sqrt {3}\pi}{8}$

Explanation

In simple cubic lattice, we have

$a = 2r$
Volume

$= \dfrac {4}{3} \pi r^3 = \dfrac {4}{3} \pi \left (\dfrac {a}{3}\right )^3 = \dfrac {1}{6} \pi a^3$

In a multi layered close-packed structure :

0%
there are twice as many tetrahedral holes as many close-packed atoms
0%
there are as many tetrahedral holes as many closed packed atoms
0%
there are twice as many octahedral holes as many close-packed atoms
0%
there are as many tetrahedral holes as many octahedral holes

Explanation

No. of octahedral holes

$=$ No. of close packed atoms
No. of Tetrahedral holes

$= 2 \times$ No. of close packed atoms

Hence, in a multi-layered close-packed structure, there are twice as many tetrahedral holes as many close-packed atoms.

An ionic compound XY forms a crystal structure in which 'Y' atoms form ccp lattice and 'X' atoms are sopositioned that each 'X'-atom touches 6'Y'-atoms. If the nearest distance between two 'X'-atoms is

$2\sqrt{2}$ ,then the density (

$g/cm^{3}$ ) of XY crystal is :

0%
$5\sqrt{6}$
0%
$\frac{20}{3}$
0%
$\frac{5}{6}$
0%
$\frac{5}{3}$

Explanation

The nearest distance between two X-atom is

$\frac{a}{\sqrt{2}}$
so

$\frac{a}{\sqrt{2}}$ =

$2\sqrt{2}$ and a = 4
d =

$\dfrac {z(molar mass)}{NA\times} a^3$ =

$\dfrac {4\times {64}}{6\times 10^{23}}\times (4\times 10^{-8} )^{3}$

$\frac{40}{6}$

$\frac{20}{3}$ g/cm3

The fraction of total volume occupied by the atoms present in a simple cubic lattice is:

0%
$\dfrac{\pi }{4}$
0%
$\dfrac{\pi }{6}$
0%
$\dfrac{\pi }{3\sqrt{2}}$
0%
$\dfrac{\pi }{4\sqrt{2}}$

Explanation

For simple cubic lattice,

Radius (r) =

$\dfrac {a}{2}$

Volume of the atom =

$\dfrac{4}{3}\times \pi \times \left ( \dfrac{a}{2} \right )^3$

Packing fraction =

$\dfrac{\dfrac{4}{3}\times \pi \times \left ( \dfrac{a}{2} \right )^3}{a^3} = \dfrac{\pi }{6}$

Hence, the correct answer is option

$\text{B}$ .

Crystal systems in which no two axial lengths are equal is/are :

0%
tetragonal
0%
orthorhombic
0%
monoclinic
0%
triclinic

Explanation

Crystal structures which has no two axial lengths equal

$( a \neq b \neq c)$ are orthorhombic, monoclinic and triclinic.

In the fluorite structure if the radius ratio is

$\left ( \sqrt{\frac{3}{2}}-1 \right )$ how many ions does each cation touch ?

0%
4 anions
0%
12 cations
0%
8 anions
0%
No cations

Which of the following statements is not correct?

0%
The number of carbon atoms in a unit cell of diamond is 4
0%
The number of Bravais lattices in which a crystal can be categorized is 14
0%
The fraction of the total volume occupied by the atoms in a primitive cell is 0.48
0%
Molecular solids are generally volatile

Which one of the following is a pseudo solid?

0%
Calcium fluoride
0%
Glass
0%
Sodium chloride
0%
All of the above

Which of the following has a regular repeated molecular pattern in three dimensional space?

0%
Solids and liquids
0%
Liquids and gases
0%
Solids
0%
Gases

Explanation

solids have a regular repeated molecular pattern in three dimensional space and this is due to their property of high intermolecular forces. But in liquids and gases the arrangement of molecular pattern it is irregular.

Hence, the correct option is

$(C)$

The edge length of the NaCl unit cell is if radius of

$Na^+$ is 97 pm and radius of it is 181 pm :

0%
556 pm
0%
278 pm
0%
139 pm
0%
120 pm

Explanation

for NaCl type structure

$2(r_-+r_+)=a$

$2(97+181) = a$

$\therefore a=2\times 278pm$

$\therefore a=556pm$
a is edge length

An element A (atomic weight =

$100$ amu) having bcc structure has unit cell edge length of

$400$ pm. Calculate the density (in g/cc) of A?

0%
$5.188$
0%
$7.188$
0%
$4.144$
0%
$10.144$

Explanation

Density(g/mol)=

$\dfrac{Z.M}{a^3.N_A}$ , where

Z: number of atoms per unit cell = 2 for BCC

M: Atomic Weight = 100 amu

a: edge cell length in cm = 400 x

$10^{-10}$ cm

$N_A$ : Avogadro Number =

$6.023\times10^{23}$

Therefore, Density =

$\dfrac{2\times100}{400\times10^{-10}\times 6.023\times10^{23}}$

= 5.188 g/mol

Hence, options A is correct.

$TiO_2$ is well known example of :

0%
triclinic system
0%
tetragonal system
0%
monoclinic system
0%
none of the above

Explanation

$TiO_2$ has tetragonal system with five plane of symmetry and five axes of symmetry.

Analysis show that nickel oxide consist of nickel ion with 96% ions having

$d^a$ configuration and 4% having

$d^7$ configuration. Which amongst the following best represents the formula of the oxide?

0%
$Ni_{1.02}O_{1.00}$
0%
$Ni_{0.96}O_{1.00}$
0%
$Ni_{0.96}O_{0.98}$
0%
$Ni_{0.98}O_{1.00}$

Explanation

Analysis show that nickel oxide consist of nickel ions with

$96\%$ having

$d^a$ and

$4\%$

$d^7$ configuration.

For

$Ni$ ,

$d^8\rightarrow Ni^{2+}, d^7\rightarrow No^{3+}$ .

Total charge of nickel.

$(0.96\times 2)+(0.04\times 3)=2.04$ .

No. of

$O^{2-}ion=\frac {2.04}{2}=1.02$ .

Formula of solid

$=NiO_{1.02}=Ni_{0.98}O$ .

Consider a cube containing n unit cells of a cubic system. A plane ABCD obtained by joining the mid points of the edges on one of its identical faces had atoms arranged as shown. Let p be the packing fraction. Choose the correct option:

0%
$n=1, p=\frac {22}{21\sqrt 2}$
0%
$n=8, p=\frac {11}{21}$
0%
$n=8, p=\frac {11}{4}$
0%
$n=1, p=\frac {11\sqrt 3}{28}$

Explanation

Cube which contains

$8$ unit cell.

By observing face

$ABCP$ , we can say that the face of the cube look like in given figure by this

$PQRS$ can be considered as a face of a unit cell which is a simple cubic arrangement. So

$n=8$ for this cube.

Since only one atom is present in simple cubic. In simple cubic

$2r=a$ .

$\therefore$ Packaging fraction

$=\cfrac{\cfrac{4}{3}\pi r^3}{a^3}=\cfrac{\cfrac{4}{3}\pi \cfrac{a^3}{8}}{a^3}=\cfrac{4}{3}\times\cfrac{22}{7}\times \cfrac{1}{8}=\cfrac{11}{21}$

$KF$ crystallizes in the

$NaCl$ type structure. If the radius of

$K^{+}$ ions is

$132$ pm and that of

$F^{-}$ ion is

$135$ pm , what is the shortest

$K-F$ distance?

0%
$267\ pm$
0%
$534\ pm$
0%
$378\ pm$
0%
$None\ of\ the\ above$

Explanation

$F^-$ ions are present in the ccp sites whereas

$K^+$ ions are present in the octahedral voids.

The shortest distance d between the anion and the cation is given by

$d = r_+ + r_- = 132 + 135$

$d = 267$ pm

Statement 1: Distance between nearest lattice points in BCC is greater than the same in FCC having same edge length.
Statement 2: FCC has greater packing efficiency than BCC.

0%
Statement-1 is true, statement-2 is true and statement-2 is the correct explanation for statement-1
0%
Statement-1 is true, statement-2 is true and statement-2 is not the correct explanation for statement-1
0%
Statement-1 is true, statement-2 is false
0%
Statement-1 is false, statement-2 is true

Explanation

Lattice type	Number of lattice points/atoms per units cell	Nearest distance between lattice points	Maximum packing density	Example
Simple cubic	$\dfrac{1}{1}$	$a$	$\dfrac{\pi}{6} = 52\%$	Phosphor
Body centered cubic	$\dfrac{2}{2}$	$\dfrac{a\sqrt{3}}{2}$	$\dfrac{\pi\sqrt{3}}{8} = 86\%$	Tungsten
Face centered cubic	$\dfrac{4}{4}$	$\dfrac{a\sqrt{2}}{2}$	$\dfrac{\pi\sqrt{2}}{3} = 74\%$	Aluminum
Diamond	$\dfrac{4}{8}$	$\dfrac{a\sqrt{2}}{2}$ Nearest distance between atoms: $\dfrac{a\sqrt{3}}{4}$	$\dfrac{\pi\sqrt{3}}{16} = 34\%$	Silicon

From the above table it can be seen that distance between BCC lattice points is greater than distance between FCC lattice points.
Also the packing efficiency of FCC is more than BCC but it is not the explanation of the previous statement.

The interstitial hole is called tetrahedral because :

0%
it is formed by four spheres
0%
partly same and party different
0%
it is formed by four spheres, the centres of which from a regular tetrahedron
0%
none of the above

If edge length of the cube is

$3.60A^o$ , then radius of carbon atom is:

0%
$0.78A^o$
0%
$0.92 A^o$
0%
$0.64 A^o$
0%
$0.35 A^o$

Explanation

Let the diameter of Carbon atom be

$d$

${4d}_c\quad sin\left(\cfrac {109}{2}\right)=\sqrt {2}a$

${4d}_c\quad sin(54^{\circ}44')=\sqrt {2}a$

$4\times d_c\times 0.8164=\sqrt {2}\times 3.6$

$d_c=1.56\mathring A$

$r_c=\cfrac {d_c}{2}=0.78\mathring A$

Calculate the density of diamond from the fact that it has a face-centered cubic structure with two atoms per lattice point and unit cell edge length of

$3.569 \times 10 ^{-8}$ cm.

0%
3.509 g/cm $^{3}$
0%
7.012 g/cm $^{3}$
0%
10.12 g/cm $^{3}$
0%
None of the above

Explanation

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

The lattice points in the FCC lattice are at the corners of the cube and face center.

Since it is given in the question that at each lattice point there are

$2$ atoms,

So, the

$z$ will be-

$Z=2\times 8\times \dfrac{1}{8}+2\times 6\times \dfrac{1}{2}$

$Z=8$

$M=12\ g/mol$ for carbon,

$N_A=6.023\times 10^{23}/mol$

$a=3.569\times 10^{-8},\quad a^3=(3.569\times 10^{-8})\ cm^{3}$

$a^3=4.546\times 10^{-23}\ cm^3$

$\rho=\dfrac{8\times 12}{6.023\times 10^{23}\times 4.546\times 10^{-23}}=3.506\ g/cm^3$

$\rho=3.5\ g/cm^3$

Hence, the correct option is A.

A unit cell of sodium chloride has four formula units. The edge of length of the unit cell is

$0.564$ nm. What is the density (in g/cc) of sodium chloride?

0%
$1$
0%
$2.16$
0%
$4.32$
0%
None of the above

If a is the lattice parameter, the volume of an fcc crystal having

$N$ atoms is :

0%
$\displaystyle \frac{Na^3}{4}$
0%
$\displaystyle \frac{Na^3}{2}$
0%
$\displaystyle Na^3$
0%
$\displaystyle 2Na^2$

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The Solid State Quiz 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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