Home
Class 12
CHEMISTRY
An element A crystallises in fcc structu...

An element A crystallises in fcc structure. 200 g of this element has `4.12xx10^(24)` atoms. If the density of A is `7.2"g cm"^(-3)`, calculate the edge length of the unit cell.

Text Solution

Verified by Experts

We know that `rho=(ZxxM)/(a^(3)xxN_(0)xx10^(-30))or a^(3)=(ZxxM)/(rhoxxN_(0)xx10^(-30))`
According to available data,
No. of atoms in the fcc unit cell (Z) = 4
Density of unit cell `(rho)=7.2"g cm"^(-3)`
Atomic mass (M) of the element may be calculated as follows :
`4.12xx10^(24)` atoms of the element are present in 200 g
`6.022xx10^(23)` atoms of the element are present in `=((200g))/(4.12xx10^(24))xx6.022xx10^(23)=29.23g`
Atomic mass (M) of the element = `29.23"g mol"^(-1)`
Avogadro's number `(N_(0))=6.022xx10^(23)mol^(-1)`
By substituting the values, `a^(3)=(4xx("29.23 g mol"^(-1)))/((7.2"g cm")^(-3)xx(6.022xx10^(23)"mol"^(-1))xx(10^(-30)"cm"^(3)))`
`=26.97xx10^(6)("pm")^(3)" "(because"value of a is in pm")`
Edge Length (a) =`[26.97xx10^(6)("pm")^(3)]^(1//3)=2.999xx10^(2)"pm"=299.9"pm"`
Promotional Banner

Similar Questions

Explore conceptually related problems

The density of a face centred cubic element (atomic mass = 40 ) is 4.25 gm cm^(-3) , calculate the edge length of the unit cell.

An element crystallises in simple cubic structure. Its density is 8g//cm^(3) and its 200g contains 24xx10^(24) atoms. Calculate the edge length of the unit cell. (b) What is meant by 'doping' in a semiconductor?

A element crystallises in a bcc structure. The edge length of its unit cell is 288 pm. If the density of the crystal is 7.3 g cm^(-3) , what is the atomic mass of the element ?

Calcium fluoride (CaF_(2)) crystallises in a cubic crystal structure. Unit cell of this structure contains four Ca^(2+) ions and eight F^(-) ions. If density of CaF_(2) is 3.2g*cm^(-3) , then calculate the edge length of the unit cell in pm.

An element crystallises in a face-centred cubic lattice. The distance between the nearest neighbours in the unit cell is 282.8pm. Calculate the edge length of the unit cell, and the radius of an atom of the element.

Silver crystallises in fcc lattice. If edge length of the cell is 4.07xx10^(-8) cm and density is 10.5g*cm^(-3) , calculate the atomic mass of silver.

The density of a metal (atomic mass = 60.2) with face centred cubic lattice is 6.25 gm cm^-3 . Calculate the edge length of the unit cell.

A metal with a molar mass of 75g*mol^(-1) crystallises in a cubic lattice structure with unit cells of an edge length of 5Å. If the density of the metal is 2g*cm^(-3) , calculate the radius of metal atom.

An element crystallizes in a body centred cubic lattice. The edge length of the unit cell is 200 pm and the density of the element is "5.0 g cm"^(-3) . Calculate the number of atoms in 100 g of this element.

An element crystallizes in f.c.c. lattice with edge length of 400 pm. The density of the element is 7 g cm^(-3) . How many atoms are present in 280 g of the element ?