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?

you want to make it 3 times stronger? Reduce two parts water, evap 2 liters.

you have .2*3= .6molesKOH. dissolved in 1 liter, it is .6M.

What percentage would that be?

To determine the percent of water that should be evaporated, we need to compare the initial and final concentrations of the KOH solution and relate them to the amount of water.

Let's calculate the initial amount of KOH in the 0.2 M KOH solution:

Initial Amount of KOH = Concentration × Volume
Initial Amount of KOH = 0.2 M × 3 L
Initial Amount of KOH = 0.6 moles

Now, let's calculate the final amount of KOH in the 0.6 M KOH solution:

Final Amount of KOH = Concentration × Volume
Final Amount of KOH = 0.6 M × 3 L
Final Amount of KOH = 1.8 moles

The difference in the amount of KOH between the initial and final solutions is:

KOH Difference = Final Amount of KOH - Initial Amount of KOH
KOH Difference = 1.8 moles - 0.6 moles
KOH Difference = 1.2 moles

Since KOH does not evaporate during the process, the change in the amount is due to the removal of water.

To find the amount of water that needs to be evaporated, we can use the fact that 1 mole of KOH is obtained from 1 mole of water in the reaction:

Amount of Water = KOH Difference
Amount of Water = 1.2 moles

Now we can calculate the initial amount of water in the 0.2 M KOH solution:

Initial Amount of Water = Total Amount of Solution - Initial Amount of KOH
Initial Amount of Water = 3 L - 0.6 moles
Initial Amount of Water = 2.4 L

To determine the percentage of water that needs to be evaporated, we need to compare the amount of water that needs to be evaporated to the initial amount of water:

Percentage of Water to be Evaporated = (Amount of Water to be Evaporated / Initial Amount of Water) × 100
Percentage of Water to be Evaporated = (1.2 moles / 2.4 L) × 100
Percentage of Water to be Evaporated = 50%

Therefore, you would need to evaporate 50% of the water from the initial 0.2 M KOH solution to obtain a 0.6 M KOH solution.

To calculate the percent of water that should be evaporated, you first need to determine the initial and final concentrations of KOH.

Given:
Initial concentration of KOH (C1) = 0.2 M
Final concentration of KOH (C2) = 0.6 M

To find the initial moles of KOH (n1), you can use the formula:

n1 = C1 * V

where:
C1 is the initial concentration of KOH (0.2 M)
V is the volume of the solution (3 L)

n1 = 0.2 M * 3 L = 0.6 moles

Now, suppose x moles of water are evaporated. The moles of KOH in the final solution (n2) will be the same as the initial moles (n1) since KOH does not evaporate:

n2 = n1 = 0.6 moles

To find the final volume of the solution (V2), you can rearrange the formula as follows:

V2 = n2 / C2

where:
C2 is the final concentration of KOH (0.6 M)
n2 is the moles of KOH (0.6 moles)

V2 = 0.6 moles / 0.6 M = 1 L

Finally, to calculate the volume of water evaporated (V_w), you subtract the final volume of the solution (V2) from the initial volume of the solution (V1):

V_w = V1 - V2
= 3 L - 1 L
= 2 L

To find the percent of water evaporated, you divide the volume of water evaporated (V_w) by the initial volume of the solution (V1) and multiply by 100:

Percent of water evaporated = (V_w / V1) * 100
= (2 L / 3 L) * 100
= 66.67%

Therefore, to obtain a 0.6 M KOH solution from a 0.2 M KOH solution, you would need to evaporate approximately 66.67% of the water.