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What is the new concentration in % (w/v) of a KOH solution made by diluting 110 mL of a 6 % (w/v) KOH solution to 330 mL? A. 2 % B. 1% C. 6% D. 12 % E. 18 %

What is the new concentration in % (w/v) of a KOH solution made by diluting 110 mL of a 6 % (w/v) KOH solution to 330 mL? A. 2 % B. 1% C. 6% D. 12 % E. 18 %

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question:** What is the new concentration in % (w/v) of a KOH solution made by diluting 110 mL of a 6% (w/v) KOH solution to 330 mL? **Options:** - A. 2% - B. 1% - C. 6% - D. 12% - E. 18% **Explanation:** To find the new concentration after dilution, use the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the initial concentration (6%), - \( V_1 \) is the initial volume (110 mL), - \( C_2 \) is the final concentration, - \( V_2 \) is the final volume (330 mL). Calculating for \( C_2 \): \[ 6\% \times 110\, \text{mL} = C_2 \times 330\, \text{mL} \] Solve for \( C_2 \): \[ C_2 = \frac{6\% \times 110\, \text{mL}}{330\, \text{mL}} = 2\% \] The correct answer is **A. 2%**.
Transcribed Image Text:**Question:** What is the new concentration in % (w/v) of a KOH solution made by diluting 110 mL of a 6% (w/v) KOH solution to 330 mL? **Options:** - A. 2% - B. 1% - C. 6% - D. 12% - E. 18% **Explanation:** To find the new concentration after dilution, use the formula: \[ C_1V_1 = C_2V_2 \] Where: - \( C_1 \) is the initial concentration (6%), - \( V_1 \) is the initial volume (110 mL), - \( C_2 \) is the final concentration, - \( V_2 \) is the final volume (330 mL). Calculating for \( C_2 \): \[ 6\% \times 110\, \text{mL} = C_2 \times 330\, \text{mL} \] Solve for \( C_2 \): \[ C_2 = \frac{6\% \times 110\, \text{mL}}{330\, \text{mL}} = 2\% \] The correct answer is **A. 2%**.
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