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Niobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm^(-3), calculate atomic radius of niobium, given its atomic mass 93 u.

Answer»

SOLUTION :`a^(3) = (M xx Z)/(d xx N_(0) xx 10^(-30))`
`a^(3) = (93 G mol^(-1) xx 2)/(8.55 g cm^(-3) xx 6.02 xx 10^(23) mol^(-1) xx 10^(-30))`
`a^(3) = 3.61 xx 10^(7) = 36.1 xx 10^(6)`
`a = (36.1)^(1//3) xx 10^(2)` pm
`= 3.304 xx 10^(2)` pm = 330.4 pm
`[x = (36.1)^(1//3), log x = 1//3 log 36.1`
`= 1//3 xx 1.5575 = 0.519` x
= antilog 0.519 = 3.305]
For body-centred cubic, `r = (SQRT(3))/(4) a = 0.4333`a
`= 0.433 xx 330.4` pm = 143.1 pm


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