An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is $6.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of atoms present in 180 g of the element is ______ $\times 10^{23}$. (Nearest integer)

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Solid State

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is

6.0 g cm^{- 3}

. The number of atoms present in 180 g of the element is ______

\times 1 0^{23}

.
(Nearest integer)

Answer: 22

For BCC, Z = 2. Using density formula:

ρ = \frac{Z \times M}{a ^{3} \times N _{A}}

, where

a = 300 \times 1 0^{- 10}

cm. Solving:

6 = \frac{2 \times M}{( 300 \times 1 0 ^{- 10} ) ^{3} \times 6.022 \times 1 0 ^{23}}

. This gives M ≈ 48.6 g/mol. Number of moles in 180 g =

\frac{180}{48.6} \approx 3.7

mol. Number of atoms =

3.7 \times 6.022 \times 1 0^{23} \approx 22.3 \times 1 0^{23}

≈ 22 × 10²³.

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer