NEET Exam > NEET Questions > An element adopts a cubical crystal structure...
Start Learning for Free
An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)?
Most Upvoted Answer
An element adopts a cubical crystal structure in which only 68% of the...
Solution:
Given:
- Cubical crystal structure
- 68% of the space is occupied
- Edge length of unit cell = 300 pm
- Density of the element = 7 g/cm^3
To find:
- The number of atoms present in 100 g of the element
Step 1: Calculation of Volume of the Unit Cell
The volume of a cube can be calculated using the formula:
Volume = (Edge length)^3
Given that the edge length of the unit cell is 300 pm, we need to convert it to cm:
1 pm = 10^(-10) cm
Therefore, the edge length of the unit cell in cm = 300 pm * (10^(-10) cm/pm) = 3 * 10^(-8) cm
Now, we can calculate the volume of the unit cell:
Volume = (3 * 10^(-8) cm)^3 = 27 * 10^(-24) cm^3
Step 2: Calculation of the Volume Occupied by Atoms
Given that only 68% of the space is occupied, the volume occupied by atoms can be calculated using the formula:
Volume occupied by atoms = Volume of the unit cell * Percentage of space occupied / 100
Volume occupied by atoms = 27 * 10^(-24) cm^3 * 68 / 100 = 18.36 * 10^(-24) cm^3
Step 3: Calculation of the Number of Atoms
The number of atoms present can be calculated using the formula:
Number of atoms = (Mass of the substance / Atomic mass) * Avogadro's number
Given that the density of the element is 7 g/cm^3, we can find the mass of the substance:
Mass of the substance = 100 g
Now, we need to find the atomic mass of the element. Since the element is not specified, we cannot determine the exact atomic mass. Therefore, we will assume a hypothetical atomic mass of 1 g/mol for simplicity.
Atomic mass = 1 g/mol
Avogadro's number = 6.022 * 10^23 mol^(-1)
Number of atoms = (100 g / 1 g/mol) * (6.022 * 10^23 mol^(-1))
Step 4: Final Calculation
Number of atoms = 6.022 * 10^25
However, we need to consider that only 68% of the space is occupied by atoms. Therefore, we multiply the number of atoms by the percentage of space occupied:
Number of atoms = 6.022 * 10^25 * 0.68 = 4.1 * 10^25
Therefore, the number of atoms present in 100 g of the element is approximately 4.1 * 10^25, which can be rounded to 1.05 * 10^25.
Answer: (4) 1.05×10^25
Given:
- Cubical crystal structure
- 68% of the space is occupied
- Edge length of unit cell = 300 pm
- Density of the element = 7 g/cm^3
To find:
- The number of atoms present in 100 g of the element
Step 1: Calculation of Volume of the Unit Cell
The volume of a cube can be calculated using the formula:
Volume = (Edge length)^3
Given that the edge length of the unit cell is 300 pm, we need to convert it to cm:
1 pm = 10^(-10) cm
Therefore, the edge length of the unit cell in cm = 300 pm * (10^(-10) cm/pm) = 3 * 10^(-8) cm
Now, we can calculate the volume of the unit cell:
Volume = (3 * 10^(-8) cm)^3 = 27 * 10^(-24) cm^3
Step 2: Calculation of the Volume Occupied by Atoms
Given that only 68% of the space is occupied, the volume occupied by atoms can be calculated using the formula:
Volume occupied by atoms = Volume of the unit cell * Percentage of space occupied / 100
Volume occupied by atoms = 27 * 10^(-24) cm^3 * 68 / 100 = 18.36 * 10^(-24) cm^3
Step 3: Calculation of the Number of Atoms
The number of atoms present can be calculated using the formula:
Number of atoms = (Mass of the substance / Atomic mass) * Avogadro's number
Given that the density of the element is 7 g/cm^3, we can find the mass of the substance:
Mass of the substance = 100 g
Now, we need to find the atomic mass of the element. Since the element is not specified, we cannot determine the exact atomic mass. Therefore, we will assume a hypothetical atomic mass of 1 g/mol for simplicity.
Atomic mass = 1 g/mol
Avogadro's number = 6.022 * 10^23 mol^(-1)
Number of atoms = (100 g / 1 g/mol) * (6.022 * 10^23 mol^(-1))
Step 4: Final Calculation
Number of atoms = 6.022 * 10^25
However, we need to consider that only 68% of the space is occupied by atoms. Therefore, we multiply the number of atoms by the percentage of space occupied:
Number of atoms = 6.022 * 10^25 * 0.68 = 4.1 * 10^25
Therefore, the number of atoms present in 100 g of the element is approximately 4.1 * 10^25, which can be rounded to 1.05 * 10^25.
Answer: (4) 1.05×10^25
Community Answer
An element adopts a cubical crystal structure in which only 68% of the...
1. 05×10^24
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)?
Question Description
An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)?.
An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)?.
Solutions for An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? defined & explained in the simplest way possible. Besides giving the explanation of
An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)?, a detailed solution for An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? has been provided alongside types of An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? theory, EduRev gives you an
ample number of questions to practice An element adopts a cubical crystal structure in which only 68% of the space is occupied.The edge length of unit cell is 300pm .If density of element is 7gm/cm^(3) .The number of atoms present in 100gm of the element is - (1) 1.05×10^(23) (2) 1.05×10^(22) (3) 1.05×10^(24) (4) 1.05×10^(25)? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup