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By X-ray diffraction it is found that nickel crystals are face-centred cubic.The edge of the unit cell is $3.52A^{\large\circ}$.The atomic mass of nickel is $58.7$ and its density is $8.94g/cm^3$.Calculate Avogadro's number from the data.

$\begin{array}{1 1}6.023\times 10^{23}\\5.023\times 10^{23}\\4.023\times 10^{23}\\3.023\times 10^{23}\end{array} $

1 Answer

Number of atoms in nickel unit cell =4
Let Avogadro's number =N
Thus the mass of one Ni atom =$\large\frac{58.7}{N}$
The mass of one unit cell =$4\big(\large\frac{58.7}{N}\big)$$g$
The volume of one unit cell $=(3.52\times 10^{-8})^3cm^3$
The density of nickel cell=$\large\frac{4(58.7/N)}{(3.52\times 10^{-8})^3}=$$8.94$
$\therefore N=6.023\times 10^{23}$
answered Aug 1, 2014 by sreemathi.v

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