Home
Class 12
CHEMISTRY
An element 'x' (Atomic mass = 40 g mol^(...

An element 'x' (Atomic mass = 40 g `mol^(-1)`) having f.c.c. structure has unit cell edge length of 400 pm. Calculate the density of 'x' and the number of unit cells in 4 g of 'x'

Text Solution

AI Generated Solution

To solve the problem, we need to calculate the density of the element 'x' and the number of unit cells in 4 g of 'x'. Let's break this down step by step. ### Step 1: Calculate the Density of 'x' The formula for density (\(d\)) is given by: \[ d = \frac{Z \cdot M}{N_A \cdot a^3} ...
Promotional Banner

Similar Questions

Explore conceptually related problems

An element 'X' (At. Mass = 40 g mol^(-1) ) having f.c.c structure has unit cell edge length of 400 pm . Calculate the density of 'X' and the number of unit cells in 4 g of 'X' . (N_(A) = 6.022 xx 10^(23) mol^(-1)) .

An element A (Atomic weight =100 ) having b.c.c. structure has unit cell edge length 400 pm. Calculate the density of A, number of unit cells and number of atoms in 10 g of A.

An element (At. Mass =50 g/mol) having fcc structure has unit cell edge length 400 pm. The density of element is

An element (atomic mass = 100 g//mol ) having bcc structure has unit cell edge 400 pm .Them density of the element is

An element (atomic mass = 100 g//mol ) having bcc structure has unit cell edge 400 pm .Them density of the element is

An element X (At,wt = 80 g//mol) having fcc structure, calculate the number of unit cells in 8g of X

An element of atomic mass 40 occurs in fcc structure with a cell edge of 540 pm. Calculate the Avogadro's number if density is 1.7gm//cm^(3) .

If the unit cell edge length of NaCl crystal is 600 pm, then the density will be

An element of atomic mass 90 occurs in fce structure with cell edge of 500 pm. Calculate the Avogadro's number if the density is 4.2g cm^(-3)

CsCl has bc c arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl .