Knowing the edge length of a cubic crystal of an element and its. Density , how will your find the value of Avogadro's number ?

By X-ray diffraction it is found that nickel ("at mass"=59g "mol"^(-1)) , crystallizes in ccp. The edge length of the unit cell is 3.5 Å . If density of Ni crystal is 9.0g//cm^(3) , then value of Avogadro's number from the data is:

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

The density of KCl is 1.9893g cm^(-3) and the length of a side unit cell is 6.29082Å as determined by X- ray diffraction. Calculation the value of Avogadro's number.

For a unit cell of edge length 10 Å, atomic mass of that element is 150 g and density is 1 g cm^(-3) .Find out the atomic radius .

Edge length of unit cell of KCI is 629 pm and density is 1.989 g per cm^3 . Calculate Avogadro's number based on this X-ray diffraction data.

In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is 625g//c.c. Avogadro number = 6times10^(23) .The crystal lattice is

Sodium chloride crystallizes in face-centred cubic (f.c.c.) structure. Its density is 2.165 g cm^(-3) . If the distance between Na^+ and its nearest Cl^(-) ions is 281 pm, find out the Avogadro's number (Na = 23 g "mol"^(-), Cl = 35.5 g "mol"^(-) ).

Derive an expression for the density of a cubic crystal whose edge is 'a' pm and contains z atoms per unit cell. The atomic mass of substance may be taken as M.

The face centred cell of nickel has an edge length 352.39 pm. The density of nickel is 8.9 g/cc. Avogadro number is

An oxide of N has vapor density of 23. Find the total number of electrons in its 92 g.(N_(A) = Avogadro's number )