Silver crystallises in fcc lattice. If edge length of the cell is `4.04x10^(-8)`cm and density is `10.5g cm^(-3)`, calculate the atomic mass of silver.

Silver crystallines in fcc lattice. If edge length of the cell is 4.077xx10^-8cm and density is 10.5 g cm^-3 . Calculate the atomic mass of silver.

Silver crystallisaes in fcc lattice. If the edge length of th cell is 4xx10^(-8) cm and density is 11.2 g "cm"^(-3) , calculate the atomic mass of silver.

An element E crystallizes in boyd centred cubic structure. If the edge lengh of the cell is 1.469xx10^-10m and the denstiy is 19.3 g cm^-3 , calculate the atomic mass of this element. Also calculate the radius of an atom of the element.

An element crystallize in fcc lattice and edge length of unit cell is 400 pm. If density of unit cell is 11.2 gcm^-3 , then atmomic mass of the element is

Silver metal crystallise with a face centred cubic .lattice. The length of unit cell is found to be 4.077 xx 10^(-8) cm . Calculate the atomic radius and density of silver (Atomic mass of Ag= 108 u, NA= 6.02 x 10^(23) mol^(-1) .

The elements crystallizes in a body centered cubic lattice and the edge of the unit cell is 0.351nm. The density is 0.533g//cm^(3) . What is the atomic mas?

An elements crystallizes in a face centered cubic lattice and the edge of the unit cell is 0.559nm. The density is 3.19g//cm^(3) . What is the atomic mass?

An element crystallising in body centred cubic lattice has an edge length of 500 pm. If its density is 4gcm^-3 , the atomic mass of the element is

An element A crystallines in fcc structure. 200 g of this element has 4.12xx10^24 atoms. The density of A is 7.2 gcm^-3 . Calculate the edge length of the unit cell.

An element X (At. Wt. =24) forms FCC lattice. If the edge length of lattice is 4 xx 10^(-8) cm and the observed density is 2.4 xx 10^(3 )kg//m^(3) . Then the percentage occupancy of lattice point by element X is : (N_(A) = 6 x 10^(23))