CBSE NCERT SOLUTIONS: Q 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 0.5 g cm

Wednesday, April 10, 2013

Q 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 0.5 g cm–3, calculate the atomic mass of silver.

Q 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^–8 cm and density is 0.5 g cm^–3, calculate the atomic mass of silver.

Solution

Edge of length of cell a = 4.07x10^–8cm

Density p = 10.5 g /cm³

Number of atoms in unit cell of fcc lattice = 4

Avogadro number N_A = 6.022x10²³

Use formula

Density $p = \frac{{ZM}}{{{a^3}{N_A}}}$

Cross multiply we get

ZM = pa³N_A

Divide by Z we get

$M = \frac{{p{a^3}{N_A}}}{Z}$

Plug the value we get

$M = \frac{{10.5{{(4.07)}^3}6.022x{{10}^{23}}}}{4}$

$M = \frac{{10.5x67.41x6.022}}{4}$

$M = 107.09g{\rm{ }}mo{l^{ - 1}}$

CBSE NCERT SOLUTIONS

Wednesday, April 10, 2013

Q 11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10–8 cm and density is 0.5 g cm–3, calculate the atomic mass of silver.

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