Wednesday, 10 April 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 NA = 6.022x1023
Use formula
        Density   p = \frac{{ZM}}{{{a^3}{N_A}}}
Cross multiply we get
                        ZM = pa3NA

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}} 
                
                       

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