A student prepared a stock solution by dissolving 15.0 g of KOH in enough water to make 150.mL of solution.She then took 15.0 mL of the stock solution and diluted it with enough water to make water to make 65.0 mL of a final solution.What is the concentration of KOH for the final solution?
A)0.411 M
B)0.534 M
C)1.87 M
D)2.43 M

Question

Quizplus · Accepted Answer

The concentration of the stock solution is calculated by dividing the amount of solute by the volume of the solution. For the stock solution: $ \text{Concentration} = \frac{15.0 \, \text{g}}{150 \, \text{mL}} $. The molar mass of KOH is approximately 56.11 g/mol, so the moles of KOH in the stock solution are $ \frac{15.0 \, \text{g}}{56.11 \, \text{g/mol}} \approx 0.267 \, \text{mol} $. The concentration of the stock solution is $ \frac{0.267 \, \text{mol}}{0.150 \, \text{L}} = 1.78 \, \text{M} $. When 15.0 mL of this solution is diluted to 65.0 mL, the moles of KOH remain the same, but the volume changes. The final concentration is $ \frac{0.267 \, \text{mol}}{0.065 \, \text{L}} = 4.11 \, \text{M} $, but this calculation mistake led to an incorrect interpretation. Correctly, the dilution formula $ C_1V_1 = C_2V_2 $ should be used, where $ C_1 $ and $ V_1 $ are the concentration and volume of the initial solution, and $ C_2 $ and $ V_2 $ are the concentration and volume of the final solution. Using the correct approach, the concentration of the final solution is found to be lower, correctly calculated as option A, 0.411 M, reflecting the dilution effect.

A Student Prepared a Stock Solution by Dissolving 15