Please answer quickly it’s an emergency

1.	Please answer quickly it’s an emergency
Answer» 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.**

Answer»

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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