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प्रश्न
An element 'X' (At. mass = 40 g mol-1) having f.c.c. the structure has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g of 'X'. (NA = 6.022 × 1023 mol-1)
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उत्तर
Unit cell edge length= 400 pm
= 400 x 100-10 cm
The volume of unit cell= a3
= (400 x 10-10 cm)3
= 64 x 10-24 cm3
Mass of unit cell = No. of atoms in the unit cell × Mass of each atom
Number of atoms in the fcc unit cell = 4
Mass of one atom = `"Atomic Mass"/"Avagadro no" = 40/(6.022 xx 10^23) g mol^(-1)`
Mass of unit cell = `(4xx40)/(6.022 xx 10^23) = 26.57 xx 10^(-23) g mol^(-1)`
Density of unit cell = `"Mass of unit cell"/"Volume of unit cell" = (26.57 xx 10^(-23))/64xx10^(-24) = 4.15 g cm^(-3)`
No of Units cells in `26.57 xx 10^23 g` = 1
No of units cells in 4g = `(1xx4)/(26.57 xx 10^23) = 0.15 xx 10^(-23)`
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