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The Solid State

Question

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An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (N_A = 6.022 × 1023 mol^–1 )

Solution

We know,

$d = \frac{Z x M}{N_{A} x a^{3}} F o r F C C Z i s 4 d = \frac{4 x 40}{6.022 x 10^{23} x (400 x 10^{- 10})^{3}} = 4.1514 g m c m^{- 3} V o l u m e o f 4 g m X i s \frac{4}{4.1514} c m^{3} = 0.96 c m^{3} V o l u m e o f 1 u n i t c e l l i s = (400 x 10^{- 10})^{3} c m^{3} = 64 x 10^{- 24} c m^{3} N u m b e r o f u n i t c e l l = \frac{0.96}{64 x 10^{- 24}} = 1.5 x 10^{22}$

Some More Questions From The Solid State Chapter

What type of solids are electrical conductors, malleable and ductile?

Give the significance of a 'lattice point'.

Name the parameters that characterize a unit cell.

Distinguish between Hexagonal and monoclinic unit cells.

Distinguish between Face-centred and end-centred unit cells.

What is the two-dimensional coordination number of a molecule in square close packed layer ?

A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3 of tetrahedral voids. What is the formula of the compound?

Which of the following lattices has the highest packing efficiency:

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