Question

calculate thenumber of oxygen ions present in 0.051 g of aluminium oxide(Al2O3)

Answer 1

Answer:

Atomic of (Al = 27u) The number of aluminum ions (Al3+) present in one molecule of aluminium oxide is 2.

Answer 2

Answer:

= 9.033 x $10^{20}$

Explanation:

1 mole of aluminium oxide ($Al_{2}O_{3}$) = (2*27) + (3 * 16) = 54 + 48 =  102 g

i.e., 102 g of $Al_{2}O_{3}$ = 6.022 * 10²³ molecules of $Al_{2}O_{3}$

Then, 0.051 g of $Al_{2}O_{3}$ contains = 3.011 * $10^{20}$ molecules of $Al_{2}O_{3}$

The number of Oxygen ions ($O^{3-}$) present in one molecule of aluminium oxide is 3.

Therefore, the number of aluminium ions (($O^{3-}$) present in 3.011 * $10^{20}$molecules (0.051 g ) of aluminium oxide ($Al_{2}O_{3}$) = 3 * 3.011 * $10^{20}$

= 9.033 x $10^{20}$