Home
Class 11
CHEMISTRY
Sodium (Na =23) crystallizen in bcc ar...

Sodium (Na =23) crystallizen in bcc arrangement with the interfacial separation between the atoms at the edge `53.6 "pm"`. The density of sodium crystal is:

A

`2.07 g//"cc"`

B

`2.46 g//"cc"`

C

`1.19 g // "cc"`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • SOLID STATE

    NARENDRA AWASTHI|Exercise Level 1 (Q.33 To Q.62)|1 Videos
  • SOLID STATE

    NARENDRA AWASTHI|Exercise Level 3 - One Or More Answers Are Correct|1 Videos
  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Exercise|196 Videos
  • STOICHIOMETRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|20 Videos

Similar Questions

Explore conceptually related problems

Sodium (Na=23) crystallizes in bc c arrangement with the interfacial separation between the atoms at the edge 53.6 pm . The density of sodium crystal is :

Sodium crystallises in bcc arrangement with the interfacial separation between the atoms at the edge length of 53 pm. The density of the solid is

Sodium crystallizes in bcc structure. If the edge length of unit cell is 4.3xx10^-s cm, the radius of Na atom is

Sodium metal crystallizes in body centred cubic lattice with the cell edge a=4.28Å. What is the radius of the sodium atom ?

An ionic solid A^(o+)B^(Θ) crystallizes as an bcc structure. The distance between cation and anion in the lattice is 338 pm . The edge length of cell is

Sodium metal crystallises in body centred cubic lattic with cell edge 427 pm .Thus radius of sodium atom is

In a metal M having BCC arrangement edge length of the unit cell is 400 pm. The atomic radius of the metal is:

Sodium metal exits in bcc unit cell .The distance between nearest sodium atoms is 0.368 nm. The edge length of the unit cell is: