Mole of electrons present in 9g …
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Sia ? 6 years, 6 months ago
Number of moles of 9 g of Al3+ ion = {tex}\frac{9}{27}{/tex}(given mass/molar mass)
Now, number of ions of Al3+ = {tex}\frac 13 \times 6.022 \times 10^{23}{/tex}= 2.007 {tex}\times{/tex} 1023(number of ions/atoms/molecules=Avogadro's number*number of moles)
Now, number of electrons in 1 ion of Al3+ = 10
So, 2.007 {tex}\times{/tex} 1023 ions will have 2.007 {tex}\times{/tex} 1023 {tex}\times{/tex} 10 = 2.007 {tex}\times{/tex} 1024 electrons.
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