Question

Calculate the number of moles of Cl− ions in 1.75 L of 1.0×10−3 M AlCl3. How would you solve this?

Answer 1

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Answer 2

The number of moles of chloride ions in the solution are 0.00525 moles

Explanation:

To calculate the number of moles from given molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of aluminium chloride = $1.0\times 10^{-3}M=0.001M$

Volume of solution = 1.75 L

Putting values in above equation, we get:

$0.001mol/L=\frac{\text{Moles of aluminium chloride}}{1.75L}\\\\\text{Moles of aluminium chloride}=(0.001mol/L\times 1.75L}=0.00175mol$

1 mole of aluminium chloride produces 1 mole of aluminium ions and 3 moles of chloride ions

Moles of chloride ions = (3 × 0.00175) = 0.00525 moles

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