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An element has bcc structure with a cell edge of 208 pm. the density of the element is 7.2 gcm^(-3). How many atoms are present in 208g of the element.

Answer»

Solution :An element has bcc STRUCTURE with a cell edge of 288 pm.
The density of the element is `"7.2 gcm"^(-3)`
For the Bcc structure, n = 2
`rho=(nM)/(a^(3)N_(A))`
`"7.2 g cm"^(-3)=(2M)/((288xx10^(-10)cm)^(3)xx(6.023xx10^(23)" mol"^(-1)))`
`"7.2 g cm"^(-3)=(2M)/((2.38xx10^(-23)cm^(3))xx(6.023xx10^(23)" mol"^(-1)))` `7.2g="0.140 M mol"`
`M=(7.2g)/("0.140 mol")="51.42 g mol"^(-1)`
By mole concept, 51.42 g of the element contains `6.023xx10^(23)` atom
208 g of the element will CONTAIN `=(6.023xx10^(23)xx208)/(51.42)" atoms"`
`=24.14xx10^(23)" atoms (or) "2.417xx10^(24)" atoms"`


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