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Test: Dimensions of Unit Cells (Old NCERT) - NEET MCQ


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10 Questions MCQ Test Chemistry Class 12 - Test: Dimensions of Unit Cells (Old NCERT)

Test: Dimensions of Unit Cells (Old NCERT) for NEET 2024 is part of Chemistry Class 12 preparation. The Test: Dimensions of Unit Cells (Old NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Dimensions of Unit Cells (Old NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Dimensions of Unit Cells (Old NCERT) below.
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Test: Dimensions of Unit Cells (Old NCERT) - Question 1

 If an element crystallizes as a simple cube, what is the volume of an element provided its density is 1.5 g/cm3 and atomic mass of the element is 63?

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 1

Simple cube has 1 atom

Therefore volume of cube i.e. a3 = 6.97 x 10-23 cm3

Test: Dimensions of Unit Cells (Old NCERT) - Question 2

Mass of an atom in an unit cell is calculated by which of the following formula if M is the molar mass:

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 2

(Molar mass = M) (1 molar of atom = M 6.023×1023)( atom = M/Na)

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Test: Dimensions of Unit Cells (Old NCERT) - Question 3

 In the formula to calculate the density of a unit cell     what is z ?

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 3

The correct answer is Option B.
Mass of unit cell = number of atoms in unit cell × mass of each atom = z × m
Where, z = number of atoms in unit cell,
m = Mass of each atom
Mass of an atom can be given with the help of Avogadro number and molar mass as: 
Where, M = molar mass
NA = Avogadro’s number
Volume of unit cell, V = a3
⇒ Density of unit cell = 
⇒ Density of unit cell = 

Test: Dimensions of Unit Cells (Old NCERT) - Question 4

Volume of a body centred cubic unit cell is

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 4

In a body centered cubic unit cell ,√3a = 4r a= 4r/√3 where a = side of the cube volume = a3 = (4r/√3)3.

Test: Dimensions of Unit Cells (Old NCERT) - Question 5

Density of a unit cell is equal to mass of unit cell divided by

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 5

Volume of unit cell, V = a3

=> Density of unit cell = mass of unit cell /volume of unit cell

Test: Dimensions of Unit Cells (Old NCERT) - Question 6

 Length a in unit cell is along

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 6

The correct answer is option A
Unit cell is essentially considered a cube, so a cube has 3 axes. X,Y,Z so X represents length , Y represents breadth, Z represents height.

Test: Dimensions of Unit Cells (Old NCERT) - Question 7

The number of atoms (z) in simple cubic unit cell is

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 7

The face-centered cubic (fcc) has a coordination number of 12 and contains 4 atoms per unit cell. The body-centered cubic (bcc) has a coordination number of 8 and contains 2 atoms per unit cell. The simple cubic has a coordination number of 6 and contains 1 atom per unit cell.

Test: Dimensions of Unit Cells (Old NCERT) - Question 8

X-ray diffraction studies show that copper crystallizes in an fcc unit cell with cell edge of 3.608 x 10-8 cm. In a separate experiment, copper is determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper.

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 8

In case of fcc lattice,number of atoms per unit cell,x=4 atoms

∴  Atomic mass of copper =63.1u

Test: Dimensions of Unit Cells (Old NCERT) - Question 9

An element with bcc geometry has atomic mass 50 and edge length 290 pm. The density of unit cell will be

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 9

Length of the edge , a = 290 pm =290 x 10-10 cm 
Volume of unit cell = ( 290 x 10-10 cm )3 = 24.39 x 10-24 cm3

Since it is bcc arrangement, 
Number of atoms in the unit cell, Z = 2 
Atomic mass of the element = 50 

Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 1023

Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 1023 = 100/6.23 x 1023
Therefore , density = mass of unit cell / volume of unit cell 
= 100/6.023 x 1023 x 24.39 x 10-24 = 6.81 g cm-3

Test: Dimensions of Unit Cells (Old NCERT) - Question 10

The number of octahedral void(s) per atom present in a cubic close-packed structure is 

Detailed Solution for Test: Dimensions of Unit Cells (Old NCERT) - Question 10

The total number of octahedral void(s) per atom present in a cubic close packed structure is 4. Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centres) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent unit cells, so is the octahedral void located on it. Only 1/4 ​th of each void belongs to a particular unit cell. Thus, in cubic close packed structure, octahedral void at the body-centre of the cube is 1.
12 octahedral voids located at each edge and shared between four unit cells=12× 1/4​=3
Total number of octahedral voids =4
We know that in ccp structure, each unit cell has 4 atoms. Thus, the number of octahedral voids is equal to 4/4=1.

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