1.

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide . (Hint : The mass of an ion is the same as that of an atom of the same element , Atomic mass of Al = 27 u)

Answer»

SOLUTION :Mole of aluminium OXIDE `(Al_(2)O_(3))`
`=2x27+3x16=102g`
i.e., `102g` of `Al_(2)O_(3)=6.022x10^(23)//102x0.051` molecules
`=3.011x10^(20)` molecules of `Al_(2)O_(3)`
The number of aluminium ion `(Al^(3+))` PRESENT in one molecule of aluminium oxide is 2. Therefore the number of aluminium ions `(Al^(3+))` present in `3.011 x 10^(20)` molecules (0.051g) of aluminium oxide
`(Al_(2)O_(3))=2x3.011xx10^(20)`
`=6.022x10^(20)`


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