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Assuming that all of the AlCl3 dissociates completely in water.
molar mass of AlCl3= 27+ 3* 35.5 => 133.5 grams per mole.
As, 267 grams of AlCl3 is dissolved in water, 267/133.5 => 2 moles of AlCl3.
The equation for dissociation is: AlCl3---------> Al+3 + 3Cl-1
So, 1 part of AlCl3 gives 1 part of Al3+ ions and 3 parts of Cl-1 ions.
So, 1 part=2 moles.
Hence, 2 moles of Al3+ and 3 X 2=> 6 moles of Cl-1 are formed.
1 mole = 6.023 X 10^23 ions.
So, 2 X 6.023 X 10^23=> 12.046 X 10^23 Al3+ ions are formed.
So, 6 X 6.023 X 10^23=> 36.138 X 10^23 Cl-1 ions are formed.
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