Question

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?

Answer 1

`\large \boxed{\text{8.90 mol/L}}`

Step-by-step explanation:

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

`\begin{array}{rcl}V_(1)c_(1) & = & V_(2)c_(2)\\\text{635 mL }* c_(1) & = & \text{1130 mL} * \text{5.00 mol/L}\\635 c_(1)&=& \text{5650 mol/L}\\c_(1)& = & (5650)/(635)\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}`

Answer 2

8.90 M

Step-by-step 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