JEE PYQ: Solid State Question 10
Question 10 - 2020 (05 Sep Shift 1)
A diatomic molecule $\text{X}_2$ has a body-centred cubic ($bcc$) structure with a cell edge of 300 pm. The density of the molecule is 6.17 g cm$^{-3}$. The number of molecules present in 200 g of $\text{X}_2$ is:
(Avogadro constant $(N_A) = 6 \times 10^{23}$ mol$^{-1}$)
(1) $40,N_A$
(2) $8,N_A$
(3) $4,N_A$
(4) $2,N_A$
Show Answer
Answer: (3)
Solution
For $bcc$, $Z = 2$. $d = \dfrac{Z \times M}{N_A \times a^3}$. $6.17 = \dfrac{2 \times M}{6.0 \times 10^{23} \times (3 \times 10^{-8})^3}$. $M = 50$. No. of moles $= \dfrac{200}{50} = 4$. No. of molecules $= 4,N_A$.
BETA
