Advertisement Remove all ads

An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (Density of an element = 14.44 g/cm3 ) - Chemistry

Sum

An element has a bcc structure with a unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (Density of an element = 14.44 g/cm3 )

Advertisement Remove all ads

Solution

Given: Type of unit cell is bcc.
Edge length (a) = 288 pm = 2.88 × 10–8 cm,
Mass of element (x) = 200 g

To find:
i. Number of unit cells in 200 g of element

ii. Number of atoms in 200 g of element

Formulae:
i. Number of atoms in x g of element = `"x n"/(rho "a"^3)`

ii. Number of unit cells in x g of element = `"x"/(rho "a"^3)`

Calculation:

i. For bcc unit cell, n = 2.

Using formula (i),

Number of atoms in 200 g of element

`= (200 "g" xx 2)/(14.44 "g cm"^-3 xx (2.88 xx 10^-8 "cm")^3) = 1.16 xx 10^24`

ii. Using formula (ii),

Number of unit cells in 200 g element

`= (200 "g")/(14.44 "g cm"^-3 xx (2.88 xx 10^-8 "cm")^3) = 5.80 xx 10^23`

∴ Number of atoms in 200 g element is 1.16 × 1024.

∴ Number of unit cells in 200 g element is 5.80 ×1023.

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 1 Solid State
Exercise | Q 10 | Page 27
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×