calculate the number of ions present in 0.051g of aluminium oxide
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6.8×10^8 Its too long to calculate
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1 mole of aluminium oxide (Al2O3) = 2*27 + 3 * 16
= 102 g
i.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains =6.022×10^23 × 0.051 ÷ 102.
= 3.011 * 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 * 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 * 3.011 * 1020
= 6.022 * 1020
= 102 g
i.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains =6.022×10^23 × 0.051 ÷ 102.
= 3.011 * 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 * 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 * 3.011 * 1020
= 6.022 * 1020
u r right
thanks
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