11.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 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

=initial molarity

=initial volume in liters

=final molarity

=final volume in liters

Solution:

using the dilution formula:

= 0.400 M

Thus, The concentration of the original solution was 0.400 M.

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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

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890