Question

When dissolved in water, AlCl3 is a strong electrolyte. How many moles of solute particles are present in 272 mL solution of 0.242 M AlCl3 ?

Answer 1

Answer: Moles of solute particles are present in 272 mL solution of 0.242 M $AlCl_3$ are 0.0658.

Explanation:

Molarity is defined as the number of moles of solute dissolved per Liter of the solution.

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of $AlCl_3$ solution = 0.242 M

Volume of solution = 272 mL

Putting values in equation 1, we get:

$0.242M=\frac{\text{Moles of}AlCl_3\times 1000}{272ml}\\\\\text{Moles of AlCl_3}=\frac{0.242mol/L\times 272}{1000}=0.0658mol$

Thus moles of solute particles are present in 272 mL solution of 0.242 M $AlCl_3$ are 0.0658.