1.

How many defects exists in the arrangement of constituent particles of 7.45 g KCl ? [K = 39, Cl = 35.5 g/mole]

Answer»

`10 xx 10^(23)`
`1 xx 10^6`
`1.0 xx 10^(-6)`
`10 xx 10^4`

SOLUTION :NUMBER of moles of KCL `= (7 : 45)/(74.5) = 0.1`
1 mol KCl = `10^6` Schottky pairs
`therefore` 0.1 mol KCl = `10^5` Schottky pairs


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