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  • A diatomic molecule X2 has a body-centred cubic (bcc) structure with a cell edge of 300 pm. The density of the molecule is <img alt="6.17\; g\; cm^{-3}" src="http://entrancecorner.oncodecogs.com/gif.latex?6.17%5C%3B%20g%5C%3B%20cm%5E%

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 X2 is : (Avogadro constant (N_{A})=6\times 10^{23}mol^{-1})
Option: 1 40\; N_{A}  
Option: 2 8\; N_{A}
Option: 3 4\; N_{A}
Option: 4 2\; N_{A}

Answers (1)

best_answer

Asking : The number of molecules present in 200 g of X2 ?

1 mole = 6 X 1023 Molecules. = NA Molecules

Mole  = mass/ Molar mass

Given mass = 200 g

Now need to find Molar mass.

 

We know this formula For Molar mass,

\text { Density of unit cell }=\frac{\mathrm{Z} \times \mathrm{M}}{\mathrm{N_A }\times \mathrm{a}^{3}}

Z = Number of atoms 

M = Molar mass

a = Cell edge length

NA  = Avogadro constant (NA ) = 6 X 1023 mol–1

 

A diatomic molecule X2 has a body-centered cubic (bcc) structure with a cell edge of 300 pm. The density of the molecule is 6.17 g cm–3.

So, 

Z = 2 for BCC

a = 300 pm = 300 X 10–10 cm

d = 6.17 g cm–3 

Now,

\textup{d = 6.17} \mathrm{~g} / \mathrm{cm}^{3}=\frac{2 \times \mathrm{M}}{6 \times 10^{23} \times\left(300 \times 10^{-10}\right)^{3}}

\mathrm{M}=\frac{6.17 \times 6 \times 9 \times 3 \times 10^{-1}}{2}

\mathrm{M}=81 \times 6.17 \times 10^{-1}

\mathrm{M}=49.97 \mathrm{~g} / \mathrm{mol}

 

Hence ,\mathrm{ Mole = \frac{200g}{49.97 g/mol}}\approx \textup{4 mol}

So, 4 mole = 4 X 6 X 1023 Molecules. = 4NA Molecules

Therefore, the correct option is (3).

Posted by

Kuldeep Maurya

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