# 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 )

#### 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.

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#### APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 1 Solid State
Exercise | Q 10 | Page 27