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Lattice Energy

An element X(...

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An element X(atomic weight=24gm/mol)forms a face centered cubic lattice.If the edge length of the lattice is 4×10^{-8}cm and the observed density is 2.40×10^3kg/m^3,then the percentage occupancy of lattice points by element X is(Use N_{A=}6×10^{23}).Explain the reason for the same also provide steps in the solution 1)96 2)98 3)99.9 4)Non

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X

= 24 g m / m o l

) forms a face centered cubic lattice. If the edge length of the lattice is

4

10^{- 8}

c m

and the observed density is

2.40

10^{3}

k g m^{3}

then the percentage occupancy of lattice points by element

X

N_{A}

=

6

10^{23}

Q. An element X (atomic weight

= 24

g/mol) forms a face centered cubic lattice. If the edge length of the lattice is

4 \times 10^{- 8}

cm and the observed density is

2.40 \times 10^{3}

^{- 3}

, then the percentage occupancy of lattice points by element X is : (Use

N_{A} = 6 \times 10^{23}

Q. An element X (atomic weight

= 24

amu) forms a face-centred cubic lattice. If the edge length of the lattice is

4 \times 10^{- 8}

cm and the observed density is

2.40 \times 10^{3}

^{- 3}

, then the percentage occupancy of lattice points by element X is: (use

N_{A} = 6 \times 10^{23}

)

Q. An element crystallises in a face-centered cubic lattice having density of

5 g / c m^{3}

and edge length of the side of unit cell is

300 p m

. Calculate mass of the element containing

3 \times 10^{24}

N_{A} = 6 \times 10^{23}

Q. An element crystalline in body centered cubic lattice has an edge of 500 pm. If its density is 4 g

c m^{- 3}

, the atomic mass of the element (in g

m o l^{- 1}

N_{A} = 6 \times 10^{23}

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