.A metal crystallises in fcc lattice.If edge length of the cell is `4.07×10^−8 cm` and density is `10.5gcm^−3` . Calculate the atomic mass of metal.

Silver crystallizes in fcc lattic. If the edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 g cm^(-3) . Calculate the atomic mass of silver.

Silver crystallizes in fcc lattic. If the edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 g cm^(-3) . Calculate the atomic mass of silver.

Silver crystallises in a fcc lattice. The edge length of its unit is 4.077xx10^(-8)cm and its density is 10.5g cm^(-3) . Claculate on this basis of the atomic mass of silver (N_(A)=6.02xx10^(23) "mol"^(-1))

Silver crystalilises in a fcc lattice. The edge length of its unit is 4.077xx10^(-8)xxcm and its density is 10.5g cm^(-3) . Claculate on this basis the atomic mass of silver (N_(A)=6.02xx10^(23) "mol"^(-1))

Silver crystallises in fcc latice. If edge length of the unit cell is 4.077xx10^(-8)cm , then calculate the radius of silver atom.

A metal crystallizes with a body-centred cubic lattice. The length of the unit cell edge is found to be 265 pm. Calculate the atomic radius.

A metal has an fcc latticed.The edge length of the unit cell is 404 pm .The density of the metal is 2.72 g//cm^(-3) .The molar mass of the metal is (N_(A) Avogadro's constant = 6.2 xx 10^(23) mol^(-1))

A metal has a fcc lattice.The edge length of the unit cell is 404 pm ,the density of the metal is 2.72g cm^(-3) . The molar mass of the metal is (N_(A) , Avorgadro's constant =6.02xx10^(23)mol^(-1))

A metal has a fcc lattice.The edge length of the unit cell is 404 pm ,the density of the metal is 2.72g cm^(-3) . The molar mass of the metal is (N_(A) , Avorgadro's constant =6.02xx10^(23)mol^(-1))

Chromium metal crystallizes with a body-centred cubic lattice. The edge length of the unit cell is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g cm^(-3) ? (atomic mass of Cr = 52.99)