1.

Find the charge of 27 g of Al^(3+)ions in coulombs.

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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