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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) - CBSE (Science) Class 12 - Chemistry

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Question

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)

Solution

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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Solution 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) Concept: Number of Atoms in a Unit Cell.
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