Respuesta :
Answer:
The concentration of the original solution was 0.400 M
Explanation:
Step 1: Data given
Volume of the aqueous ethanol solution = 400 mL = 0.400 L
The solution was diluted to 4.00 L
The concentration of the diluted solution is 0.0400 M
Step 2: Calculate the concentration of the original solution
We use the dilution equation:
C1V1 = C2V2
⇒with C1 = the concentration of the initial solution = TO BE DETERMINED
⇒with V1 = the volume of the initial solution= 0.400 L
⇒with C2 = the concentration of the final solution= 0.0400 M
⇒with V2 = the volume of the final solution= 4.00 L
C1* 0.400 L = 0.0400 M * 4.00 L
C1 = (0.0400 M * 4.00L) / 0.400 L
C1 = 0.400 M
The concentration of the original solution was 0.400 M
An aqueous ethanol solution (400 mL) was diluted to 4.00 L, giving a concentration of 0.0400 M. The concentration of the original solution was 0.400 M.
Given:
ethanol solution V₁ = 400 ml
concentration M₂ = 0.0400 M
volume of diluted ethanol V₂ = 4.00 L = 4000 ml
concentration of original solution = M₁ = ?
Formula used:
Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution.
in this case, we need to use the dilution formula
M₁V1 = M₂V₂
[tex]M_{1}[/tex] = initial molarity
[tex]V_{1}[/tex] = initial volume in liters
[tex]M_{2}[/tex] = final molarity
[tex]V_{2}[/tex] = final volume in liters
Solution:
using the dilution formula:
[tex]M_{1} \times 400 = 0.04 \times 4000\\\\M_{1} = \frac{0.04 \times 4000}{400}[/tex]
= 0.400 M
Thus, The concentration of the original solution was 0.400 M.
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