Question

Select the container from the figure (Figure 1) that represents the dilution of a 4 % (m/v) KCl solution to each of the following: Figure1 of 1 There is a diagram showing several containers. One container is filled with 4 percent of mass to volume solution of NaCl. Container 1 is filled with a solution in which volume is two times less than the volume of NaCl solution. Container 2 is filled with a solution of a volume two times larger compared to the NaCl solution. Container 3 is filled with a solution of a volume two times larger than the volume of the solution in container 2. Part A 2 % (m/v) KCl

Answer 1

From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.

The correct responses are;

• Part A: Container 2

• Part B: Container 3

Reasons:

The given parameters are;

The concentration of the KCl solution = 4% m/v

Taking the solution as solution of KCl

The volume of the solution in container 1 = Two times less than the volume of KCl solution.

$V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}$

The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.

$V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}$

The volume of the solution in container 3 = Two times larger than the container two solution volume.

$V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}$

Therefore;

$V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}$

Part A Required:

a. To select the container that represent the dilution of the 4% solution to 2%

Solution:

The dilution formula is; C₁·V₁ = C₂·V₂

Therefore;

$\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}$

C₁ =4%, C₂ = 2%, we get;

$\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1$

The volume of the container that represents a 2% dilution is container 2

which is filled with a solution of a volume two times larger compared to the

KCl solution.

Part B:

Required:

The container diluted to a 1% m/v KCl solution.

Solution;

Using the dilution formula, we have;

C₁ = 4%, C₂ = 1%

Therefore;

$\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}$

$\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}$

The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.

Possible parts of the question are;

Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;

Part A: a 2% (m/v) KCl solution

Part B: a solution that is a 1% (m/v) KCl solution

Please see attached drawings

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