1.

Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of Iron is 7.874 g cm^(-3). Use this information to calculate Avogadro's number. (At mass of Fe = 55.845 u).

Answer»

Solution :`d = (Z xx M)/(a^(3) xx d)`
For bcc lattice Z = 2
`N_(A) = (2 xx 55.845 G mol^(-1))/((286.65 xx 10^(-10) cm)^(3) N_(A))`
`N_(A) = (2 xx 55.845 g mol^(-1))/((286.65 xx 10^(-10) cm)^(3) xx 7.874 g cm^(-3))`
`N_(A) = 6.04 xx 10^(23) mol^(-1)`


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