A 0.2 M KOH solution has a volume of 3 L.
What percent of water should be evaporated to obtain a 0.6 M KOH solution?
The answer I got is, 66.7%
looks good
To find the percentage of water that needs to be evaporated, we can compare the initial and final concentrations of the KOH solution.
Let's start by calculating the initial number of moles of KOH in the 0.2 M KOH solution. The molarity (M) is defined as moles of solute divided by the volume of the solution in liters.
Initial moles of KOH = Molarity × volume of solution = 0.2 M × 3 L = 0.6 moles
For a 0.6 M KOH solution, we want to find the final volume of the solution after some water is evaporated.
Final moles of KOH = Molarity × volume of solution
Since we want the final molarity to be 0.6 M, we can rearrange the equation to solve for the volume of the solution:
Volume of solution = Final moles of KOH / Molarity of KOH = 0.6 moles / 0.6 M = 1 L
Now, let's find the volume of water that needs to be evaporated in order to obtain a 1 L solution.
Volume of water to be evaporated = Initial volume of solution - Final volume of solution = 3 L - 1 L = 2 L
Finally, to calculate the percentage of water to be evaporated, we can use the formula:
Percentage = (Volume of water to be evaporated / Initial volume of solution) × 100
Percentage = (2 L / 3 L) × 100 ≈ 66.7%
Therefore, your answer of 66.7% is correct.