Respuesta :

maacastrobr

Answer:

105.5 mL

Explanation:

To solve this problem we can use the following equation:

  • C₁V₁=C₂V₂

Where:

  • C₁ = 0.504 M
  • V₁ = ?
  • C₂ = 0.244 M
  • V₂ = 218 mL

We input the data:

  • 0.504 M * V₁ = 0.244 M * 218 mL

And solve for V₁:

  • V₁ = 105.5 mL
Eduard22sly

The volume of 0.504 M barium nitrate solution is needed to prepare the solution is 52.77 mL

We'll begin by calculating the molarity of the diluted barium nitrate solution. This can be obtained as follow:

Ba(NO₃)₂(aq) —> Ba²⁺(aq) + 2NO₃¯(aq)

From the balanced equation above,

2 mole of NO₃¯ is present in 1 mole of Ba(NO₃)₂

Therefore,

0.244 M NO₃¯ will be present in = 0.244 / 2 = 0.122 M Ba(NO₃)₂

Finally, we shall determine the volume of 0.504 M barium nitrate needed to prepare the solution

Molarity of stock solution (M₁) = 0.504 M

Volume of diluted solution (V₂) = 218 mL

Molarity of diluted solution (M₂) = 0.122 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

0.504 × V₁ = 0.122 × 218

0.504 × V₁ = 26.596

Divide both side by 0.504

V₁ = 26.596 / 0.504

V₁ = 52.77 mL

Therefore, the volume of 0.504 M barium nitrate needed to prepare the solution is 52.77 mL

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