Question

There are ________ mol of bromide ions in 0.300 l of a 0.150 m solution of albr3.

Answer 1

The number of moles of AlBr3 = Molarity × Volume
= 0.150 × 0.3 
= 0.045 moles
1 mole of AlBr3 gives 3 moles of Br- ions
That is 
AlBr3 = Al3+ + 3Br-
Therefore, 0.045 moles of AlBr3 will yield 3 × 0.045 = 0.135 moles
Thus; they are 0.135 moles of bromide ions

Answer 2

Moles are determined as the product of the molar concentration and the volume of the solution. The number of moles of bromide ions is 0.135 mol.

What are moles?

Moles are the product of the molarity and the volume in liters. The formula for a mole is given as,

$\rm Moles = \rm Molarity \times Volume$

Given,

Molarity = 0.150 M

Volume of solution = 0.300 L

Substituting value in the above equation:

$0.150 \times 0.3 = 0.045 \;\rm moles$

Aluminum bromide dissociates as:

$\rm AlBr_{3} \rightarrow Al^{3+} + 3Br^{-}$

From the above, it can be said that 1 mole of aluminum bromide gives 3 moles of bromide ions.

Hence, 0.045 moles of aluminum bromide will give 0.135 moles of bromide ions.

Therefore, there are 0.135 moles of bromide ions in a 0.150 M solution.

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