1.

femous oxide has cubes structure and each edgeof the unit cell is 5.0 Å .Assuming of the oxide as 4.0g//cm^(3) then the number of Fe^(2+) and O^(2) inos present in each unit cell will be

Answer»

`FOUR Fe^(2+) and two O^(2-)`
`Two Fe^(2+) and four O^(2-)`
`Four Fe^(2+) and Four O^(2-)`
`Three Fe^(2+) and three O^(2-)`

Solution :Let the units of femousoxide in a unit cell `= n` molecular weight of femous oxids `(FeO) = 56 + 16 = 73 g mol^(-1)`
Weight of a unit `= (72 xx n)/(6.023 xx 10^(23))`
VOLUME of one cell `= ("length of corner")^(2)`
`= (5 Å)^(2) = 125 xx 10^(-29) cm^(3)`
`"Density" = ("wt. of cell")/("volume")`
`4.09 = (72 xx n)/(6.023 xx 10^(23) xx 125 xx 10^(-24))`
`n = (3079.2 xx 10^(-1))/(72)`
`= 42.7 xx 10 = 4.27 ~~ 4 `


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