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Volume of a sphere

Each edge of ...

Question

Each edge of a cubic unit cell is $400 pm$ long. If atomic weight of the element is $120$ and it's density is $6.25 {g/cm}^{3}$ . The crystal lattice is:
(Given: $N_{A} = 6 \times 10^{23}$ )

Body centered

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Face centered

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Primitive

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End centered

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Solution

The correct option is A Body centered
We know that, density $(d) = \frac{z M}{N_{A} (a^{3})} \Rightarrow z = \frac{d N_{A} a^{3}}{M} \Rightarrow z = \frac{6.25 {g/cm}^{3} \times 6 \times 10^{23} (400 \times 10^{- 10} cm)^{3}}{120} \Rightarrow z = 2$
$∴$ It is a body centered lattice.

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Similar questions

Each edge of a cubic unit cell is 400pm long. If atomic weight of the element is 120 and it's density is 6.25 g /cm3 . The crystal lattice is :
1 primitive
2 body centered
3 face centered
4 end centered

Q. The element crystallizes in a body centred cubic lattice and the edge of unit cell is

0.351 n m

. The density is

0.533 g / c m 3

. What is the atomic weight?

Q. If an element (atomic weight

= 50

) crystallizes in FCC lattice, with

a = 0.50 nm

, the density

({in g/cm}^{3})

of unit cell if it contains

0.25 %

Schottkey defects is: (use

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

)

Q. An element A (Atomic weight = 120) having bcc structure has unit edge length 400 pm. Identify the correct option(s). (Given:

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}

atoms.
Take

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

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Volume of a sphere

Standard IX Mathematics

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Each edge of a cubic unit cell is 400 pm long. If atomic weight of the element is 120 and it's density is 6.25 g/cm3. The crystal lattice is: (Given: NA=6×1023)

The correct option is A Body centeredWe know that, density (d)=zMNA(a3)⇒z=d NA a3M⇒z=6.25 g/cm3×6×1023(400×10−10 cm)3120⇒z=2 ∴ It is a body centered lattice.

Each edge of a cubic unit cell is $400 pm$ long. If atomic weight of the element is $120$ and it's density is $6.25 {g/cm}^{3}$ . The crystal lattice is:
(Given: $N_{A} = 6 \times 10^{23}$ )

The correct option is A Body centered
We know that, density $(d) = \frac{z M}{N_{A} (a^{3})} \Rightarrow z = \frac{d N_{A} a^{3}}{M} \Rightarrow z = \frac{6.25 {g/cm}^{3} \times 6 \times 10^{23} (400 \times 10^{- 10} cm)^{3}}{120} \Rightarrow z = 2$
$∴$ It is a body centered lattice.