Respuesta :
Answer: Moles of solute particles are present in 272 mL solution of 0.242 M [tex]AlCl_3[/tex] 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:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
Molarity of [tex]AlCl_3[/tex] solution = 0.242 M
Volume of solution = 272 mL
Putting values in equation 1, we get:
[tex]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[/tex]
Thus moles of solute particles are present in 272 mL solution of 0.242 M [tex]AlCl_3[/tex] are 0.0658.
The number of moles of solute particles present in the solution is 0.0658 mole
Stoichiometry
From the question,
We are to determine how many moles of solute particles are present.
Using the formula,
Number of moles = Concentration × Volume
From the given information,
Concentration = 0.242 M
Volume = 272 mL = 0.272 L
Putting the parameters into formula, we get
Number of moles = 0.242 × 0.272
Number of moles = 0.065824 mole
Number of moles ≅ 0.0658 mole
Hence, the number of moles of solute particles present in the solution is 0.0658 mole
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