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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)?
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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
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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)?
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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)?.
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