An element (atomic weight $= 50$ amu) crystallizes in fcc lattice with edge length $a$ equal to $0.50$ nm. What is the density (in g/cc) of unit cell if it contains $25 %$ Schottky defects?(Use $N_{A} = 6 \times 10^{23}$ )

2.0

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2.66

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3.06

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None of the above

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Solution

The correct option is D $2.0$
$25 %$ of Schottky defects means it contains $3$ atoms per unit cell instead of $4$ .
Density $(d) = \frac{n M}{V N_{A}} = \frac{3 \times 50}{((5 \times 10^{- 8})^{3} \times 6 \times 10^{23})} = 2$ g/cc

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Q. An element (at. Wt. = 50) crystallises in FCC lattice, with a = 0.50 nm. What is the density of the unit cell if it contains 0.25% Schottky defects?

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(a = Edge length of the unit cell)

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Q. An element (atomic weight = 50) crystallises in an fcc lattice, with a = 0.50 nm. The density of the unit cell (in gm/cc) if it contains 0.25% Schottky defects is

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An element (atomic weight =50 amu) crystallizes in fcc lattice with edge length a equal to 0.50 nm. What is the density (in g/cc) of unit cell if it contains 25% Schottky defects?(Use NA=6×1023)

The correct option is D 2.025% of Schottky defects means it contains 3 atoms per unit cell instead of 4.Density (d)=nMVNA=3×50((5×10−8)3×6×1023)=2 g/cc

An element (atomic weight $= 50$ amu) crystallizes in fcc lattice with edge length $a$ equal to $0.50$ nm. What is the density (in g/cc) of unit cell if it contains $25 %$ Schottky defects?(Use $N_{A} = 6 \times 10^{23}$ )

The correct option is D $2.0$
$25 %$ of Schottky defects means it contains $3$ atoms per unit cell instead of $4$ .
Density $(d) = \frac{n M}{V N_{A}} = \frac{3 \times 50}{((5 \times 10^{- 8})^{3} \times 6 \times 10^{23})} = 2$ g/cc