Q. 114.3( 34 Votes )

# Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^{–8} cm and density is 10.5 g cm^{–3}, calculate the atomic mass of silver.

Answer :

Given edge length (a) = 4.07 × 10 ^{-8} cm

Density (d) = 10.5gcm^{-1}

For fcc number of atoms per unit cell (z) = 4

Avagadro number N_{A} = 6.022 × 10^{23} _{mol-1}

We know that, Atomic mass (M)

(M)

Thus, M

⇒ M = 106.5 mol^{-1}

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PREVIOUSCalculate the efficiency of packing in case of a metal crystal forface-centred cubic (with the assumptions that atoms are touching each other).NEXTA cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

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