A diatomic molecule X2 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 X 2 is (Avogadro constant .( N A )=6 × 1023 mol -1)

Question

A diatomic molecule X2 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 X 2 is (Avogadro constant .( N A )=6 × 1023 mol -1) (A) 8 N A  (B) 40 N A  (C) 4 N A  (D) 2 N A

Tardigrade Admin · Accepted Answer

p =(2 × ( M / N A )/ a 3) ⇒ 6.17=(2 × ( M / N A )/(3 × 10-8 cm )3) ⇒ M ≃ 50 gm / mol No =( w / M ) × N A =(200/50) × N A =4 N A

Q. A diatomic molecule $X_{2}$ has a body-centred cubic (bcc) structure with a cell edge of $300 p m$ . The density of the molecule is $6.17 g c m^{- 3}$ . The number of molecules present in $200 g$ of $X_{2}$ is (Avogadro constant $(N_{A}) = 6 \times 1 0^{23} m o l^{- 1})$

$8 N_{A}$

$40 N_{A}$

$4 N_{A}$

$2 N_{A}$

$p = \frac{2 \times \frac{M}{N _{A}}}{a ^{3}}$

$\Rightarrow 6.17 = \frac{2 \times \frac{M}{N _{A}}}{( 3 \times 1 0 ^{- 8} c m ) ^{3}}$

$\Rightarrow M ≃ 50 g m / m o l$

No $= \frac{w}{M} \times N_{A} = \frac{200}{50} \times N_{A} = 4 N_{A}$

Q. A diatomic molecule X2​ has a body-centred cubic (bcc) structure with a cell edge of 300pm. The density of the molecule is 6.17gcm−3. The number of molecules present in 200g of X2​ is (Avogadro constant (NA​)=6×1023mol−1)

8NA​

40NA​

4NA​

2NA​