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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoAnswer the following question:
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)
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
Molar mass of aluminium oxide (\(Al_2O_3\)).
= (2 x 27) + (3 x 16).
= 54 + 48 = .
= (2 x 27) + (3 x 16).
= 54 + 48 = .
Therefore,
102 g of \(Al_2O_3\) contains = 2 x \(6.022\) x \(10^2{^3}\) aluminium ions
0.051 g of \(Al_2O_3\) contains = 2 x \(6.022\) x \(10^2{^3}\) / 102 x 0.051
= x \(10^2{^3}\) x 0.051 / 102
= 0.614 x \(10^2{^3}\) /
= 0.006022 x \(10^2{^3}\)
= x \(10^2{^0}\) \(Al^3{^+}\) ions.
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