Question

An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405pm. If density of the element is 2.7g/cm³. What is the nature of cubic unit cell? (Hint : find value of z)
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Answer 1

Answer:

Explanation:

$d = ZM/a^{3} N$

$Z=dNa^{3} /M$

on substituting u will get the answer approx 3.96 can be rounded to 4

so FCC      

Answer 2

$\huge\rm\underline\purple{Question :-}$

An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405pm. If density of the element is 2.7g/cm³. What is the nature of cubic unit cell?

$\huge\rm\underline\purple{Answer :-}$

$\green{\underline\bold{Given,}}$

• Molar mass of the given element,m = 27 g/mol= 0.027 kg/mol
• Edge length,a = 4.05 × 10^-10m

$\red{\underline\bold{To\:find,}}$

• Nature of the cubic unit cell = ?

$\orange{\underline\bold{Applying\: the\: relation ,}}$

$\boxed{ \sf \blue{ d = \frac{z×m}{a3×Na}}}$

where, z is the number of atoms in the unit cell and Na is the avogadro number.

$\pink{\underline\bold{Thus,}}$

z = $\frac{d×a3×Na}{m}$

✏ After calculating, we get the value of z as,

z = 4

✏ Since, the number of atoms in the unit cell is 4, the given cubic unit cell has a fcc or ccp structure.

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