1.

Calculate the number of aluminium ions present in 0.051g of aluminium oxide.

Answer»

Solution :Molar MASS of aluminium oxide `Al_(2)O_(3) = (2 xx 27) + (3 xx 16)` = 54 + 48 = 102 G
The number of molecules
`= ("given mass")/("molar mass") xx "Avogadro number"`
`= (0.051)/(102) xx 6.022 xx 10^(23)`
`= 0.003011 xx 10^(23) ~~ 3.011 10^(20)`
1 molecule of `Al_(2)O_(3)` give `2Al^(3+)` lons
`:. 0.051 g Al_(2)O_(3)` gives `2 xx 3.011 xx 10^(20)`
`= 6.022 xx 10^(20)` ions


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