1.

Silver crystallizes in ſec lattice. If edge length of the cell is 4.077 xx 10^(-8) cm and density is 10.5 g cm^(-3). Calculate the atomic mass of silver.

Answer»

Solution :`"Density" rho= (Z xx M)/(a^(3) xx N_(0))`
or`M= (rho xx a^(3) xx N_(0))/(Z)`
Z= 4(fcc lattice)
`rho= 10.5 g cm^(-3), N_(0)= 6.022 xx 10^(23), a= 4.077 xx 10^(-8)` cm
`:. M= ((10.5 xx 4.077 xx 10^(-8))^(3) xx 6.022 xx 10^(23))/(4)=107.12 "g MOL"^(-1)`


Discussion

No Comment Found

Related InterviewSolutions