To determine how much of Solution B must be added to Solution A to achieve the desired concentration of 55%, we can set up an equation based on the amounts and concentrations of the solutions.
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Calculate the amount of solute in Solution A and Solution B.
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For Solution A (5 liters at 50%): \[ \text{Amount of solute in Solution A} = 5 \text{ liters} \times 0.50 = 2.5 \text{ liters} \]
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For Solution B (15 liters at 20%): \[ \text{Amount of solute in Solution B} = 15 \text{ liters} \times 0.20 = 3.0 \text{ liters} \]
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Let \( x \) be the amount of Solution B (in liters) to be added to Solution A. When we add \( x \) liters of Solution B to Solution A, the total volume becomes: \[ 5 + x \text{ liters} \] The total amount of solute now will be: \[ 2.5 \text{ liters (from Solution A)} + 0.2x \text{ liters (from Solution B)} \] Here, \( 0.2x \) comes from the concentration of Solution B (20%).
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We want the new solution to be 55% diluted. Therefore, we set up the equation: \[ \frac{2.5 + 0.2x}{5 + x} = 0.55 \]
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Now we can solve this equation: \[ 2.5 + 0.2x = 0.55(5 + x) \] Expanding the right side: \[ 2.5 + 0.2x = 2.75 + 0.55x \]
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Rearranging gives: \[ 2.5 - 2.75 = 0.55x - 0.2x \] \[ -0.25 = 0.35x \] \[ x = \frac{-0.25}{0.35} = -\frac{25}{35} = -\frac{5}{7} \approx -0.714 \text{ liters} \]
Since we obtained a negative value for \( x \), it indicates that you cannot achieve a 55% concentration by adding any amount of Solution B; instead, you would need to remove some volume of Solution B.
Hence, this leads us to conclude that the answer is:
C. The answer is extraneous.
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