You visited us 6 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Face centered

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Primitive

No worries! We‘ve got your back. Try BYJU‘S free classes today!

End centered

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

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}

Join BYJU'S Learning Program

Explore more

Volume of a sphere

Standard IX Mathematics

Join BYJU'S Learning Program

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.