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A 635 mL NaCl solution is diluted to a volume of 1.13 L and a concentration of 5.00 M . What was the initial concentration C1?

Respuesta :

dsdrajlin

Answer:

8.90 M

Explanation:

Step 1: Given data

  • Initial concentration (C₁): ?
  • Initial volume (V₁): 635 mL = 0.635 L
  • Final concentration (C₂): 5.00 M
  • Final volume (V₂): 1.13 L

Step 2: Calculate the initial concentration

We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₁ = C₂ × V₂  / V₁

C₁ = 5.00 M × 1.13 L  / 0.635 L

C₁ = 8.90 M

znk

Answer:

[tex]\large \boxed{\text{8.90 mol/L}}[/tex]

Explanation:

We can use the dilution formula to calculate the concentration of the original solution.

[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]

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