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Find the charge of 27 g of Al^(3+)ions in coulombs. |
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Answer» Solution :One `AL^(3+)`ion has the charge of 3 protons, and a proton has the same magnitude of charge as that on an ELECTRON. No. Of moles of `Al^(3+)` ions `=("wt. In g")/(at. Wt")` `=27/27=1` No. Of `Al^(3+)` ions in 27 g = no. Of moles x Av. const `=1 xx 6.022 xx 10^(23)` Charge of 27 g of `Al^(3+)` ions =`3 xx` charge of a proton `xx` no. of `Al^(3+)` ions `=3 xx 1.602 xx 10^(-19) xx 6.022 xx 10^(23)` `=2.894 xx 10^(5)` coulombs. |
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