Question

The number of unit cells present in 100 g of a BCC
crystal with density 5 g/cm3 and edge length
equal to 100 pm is
(1) 4 x 10^22
(2) 2 x 10^25
(3 4 x 10^23
(4) 2 x 10^24​

Answer 1

2×10^22 is the answer of this question

Answer 2

Concept:

BCC unit cell: In addition to the eight atoms at the cube's eight corners, one atom is positioned inside the cube itself. Eight corners of the lattice cube each share one-eighth of an atom. The number of atoms in a BCC unit cell is 2.

Given:

Mass of BCC crystal = 100 g

Density = 5 g/cm³

Edge length = 100 pm = 100 * 10⁻¹⁰ cm = 10⁻⁸ cm

Find:

The number of unit cells present in 100 g of a BCC crystal with a density of 5 g/cm³ and edge length equal to 100 pm is?

Solution:

We know that the density can be expressed as:

Density = $\frac{Z * M}{a^{3} * N_{A} }$

where Z is the number of atoms per unit cell (BCC) = 2

M is the molar mass.

a is the edge length

By substituting the given values in this expression, we can calculate the molar mass (M) of the BCC crystal.

5 g/cm³ = $\frac{2 * M}{(10^{-8})^{3} * 6.022*10^{23} }$

M = $\frac{5 * (10^{-8})^{3} * 6.022 * 10^{23} }{2}$

M = 15.055 * 10⁻¹

M = 1.5055

Now, moles of BCC crystal can be calculated as:

Moles of BCC crystal = $\frac{Mass }{Molar mass}$ = $\frac{100}{1.5055}$ = 66.423

Therefore, the number of moles of BCC crystal in per mole = 66.423 * 6.022 * 10²³ = 399.99 * 10²³ = 400 * 10²³

Now, the number of unit cells in a BCC crystal can be calculated as:

Number of unit cells in a BCC crystal = $\frac{1}{2}$ * 400 * 10²³

= 200 * 10²³

= 2 * 10²⁵

Hence, the correct option is (2) 2 * 10²⁵.

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