A diatomic molecule $\mathrm{X}_{2}$ has a body centered cubic (bcc). The density of the molecule is $6.17 \mathrm{~g} \mathrm{~cm}^{-3}$. The number of molecules present in 200 g of $\mathrm{X}_{2}$ is: (Avogadro constant $(\mathrm{N}_{\mathrm{A}})=6 \times 10^{23} \mathrm{~mol}^{-1}$)

(D) $4 \mathrm{~N}_{\mathrm{A}}$

(A) $2 \mathrm{~N}_{\mathrm{A}}$

(B) $8 \mathrm{~N}_{\mathrm{A}}$

(C) $14 \mathrm{~N}_{\mathrm{A}}$

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Solid State

A diatomic molecule

X_{2}

has a body centered cubic (bcc). The density of the molecule is

6.17 g cm^{- 3}

. The number of molecules present in 200 g of

X_{2}

is: (Avogadro constant

(N_{A}) = 6 \times 1 0^{23} mol^{- 1}

)

Answer: (D)

A body-centred cubic cell contains

Z = 2

formula units. From

ρ = \frac{Z M}{N _{A} a ^{3}}

with

ρ = 6.17

g/cm

^{3}

and the given cell edge, the molar mass of

X X_{2}

works out to

50

g/mol. Then

200

g is

\frac{200}{50} = 4

mol, which holds

4 N_{A}

molecules.

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option