The face centered cubic cell of platinum has a length of o.392nm. Calculate the density of platinum \[(g/c{m^{3)}}\].
tomic weight of Pt=$195$
(A) 20.9
(B) 20.4
(C) 19.6
(D) 21
Answer
Verified
115.5k+ views
Hint: Platinum adopts a face centered cubic crystal structure. This is also known as cubic close packed structure. It consists of 4 equivalent metal atoms in a cubic unit cell.
Formula used:
$d = \dfrac{{Z \times M}}{{{N_a} \times {a^3}}}$
Where, Z = Number of atoms in fcc crystal = 4
M= 195 (given in question)
${N_a}$= Avogadro’s number =$6 \times {10^{23}}mo{l^{ - 1}}$
d=density
a=$0.392nm = 0.39 \times {10^{ - 7}}cm$
Complete step by step solution:
The density of a unit cell is given as the ratio of mass and volume of unit cells.The mass of a unit cell is equal to the product of number of atoms in a unit cell and the mass of each atom in the unit cell.
Mass of unit cell=number of atoms in unit cell$ \times $mass of each atom
$ = z \times m$ ( where z=number of atoms in unit cell and m=mass of each atom)
Mass of atom can be given with the help of Avogadro’s number and molar mass, this is given as:
$\dfrac{M}{{{N_A}}}$
Now, volume of unit cell, $V = {a^3}$
Therefore, the density of unit cell will be $\dfrac{m}{v} = \dfrac{{z \times m}}{{{a^3}}}$
$ = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
According to the above given formula,
We will find out the density of platinum.
Now,
$d = \dfrac{{4 \times 195}}{{6 \times {{10}^{23}} \times {{(0.392 \times {{10}^{ - 7}})}^3}}}$
$d = 21gc{m^{ - 3}}$
Therefore,
The density is $21gc{m^{ - 3}}$
Hence, option D is correct.
Note: Platinum is a silvery white metal. It is extremely resistant to tarnishing and corrosion (which makes it a noble metal). It is used in chemical industries as a catalyst for the production of nitric acid, silicon and benzene. It is one of the densest precious metals, followed by gold, mercury, lead and silver.
Formula used:
$d = \dfrac{{Z \times M}}{{{N_a} \times {a^3}}}$
Where, Z = Number of atoms in fcc crystal = 4
M= 195 (given in question)
${N_a}$= Avogadro’s number =$6 \times {10^{23}}mo{l^{ - 1}}$
d=density
a=$0.392nm = 0.39 \times {10^{ - 7}}cm$
Complete step by step solution:
The density of a unit cell is given as the ratio of mass and volume of unit cells.The mass of a unit cell is equal to the product of number of atoms in a unit cell and the mass of each atom in the unit cell.
Mass of unit cell=number of atoms in unit cell$ \times $mass of each atom
$ = z \times m$ ( where z=number of atoms in unit cell and m=mass of each atom)
Mass of atom can be given with the help of Avogadro’s number and molar mass, this is given as:
$\dfrac{M}{{{N_A}}}$
Now, volume of unit cell, $V = {a^3}$
Therefore, the density of unit cell will be $\dfrac{m}{v} = \dfrac{{z \times m}}{{{a^3}}}$
$ = \dfrac{{z \times M}}{{{a^3} \times {N_A}}}$
According to the above given formula,
We will find out the density of platinum.
Now,
$d = \dfrac{{4 \times 195}}{{6 \times {{10}^{23}} \times {{(0.392 \times {{10}^{ - 7}})}^3}}}$
$d = 21gc{m^{ - 3}}$
Therefore,
The density is $21gc{m^{ - 3}}$
Hence, option D is correct.
Note: Platinum is a silvery white metal. It is extremely resistant to tarnishing and corrosion (which makes it a noble metal). It is used in chemical industries as a catalyst for the production of nitric acid, silicon and benzene. It is one of the densest precious metals, followed by gold, mercury, lead and silver.
Recently Updated Pages
JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key
JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key
JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key
JEE Main Chemistry Exam Pattern 2025 (Revised) - Vedantu
JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key
Trending doubts
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Thiol group is present in A cystine B cysteine C methionine class 12 chemistry JEE_Main
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Current Loop as Magnetic Dipole and Its Derivation for JEE
JEE Main 2023 January 30 Shift 2 Question Paper with Answer Keys & Solutions