what maximum volume of 3 M solution can be prepared from 1 L each of 1 M KOH and 6 M KOH solutions:

1, without adding water?

2, by using water?

(1) Without adding water

We know that the Molarity equation is given by:

M1V1 + M2V2 = M3V3, where M1,M2,M3 are concentrations and V1,V2,V3 are their respective volumes.

For the first solution, M1 = 6 (6M KOH), M2 = 1 (1M KOH), V1 & V2 need to be determined.

For the resulting solution, we want M3 to be 3 (3M KOH) and V3 = V1 + V2, which needs to be maximized.

6V1 + 1V2 = 3(V1+V2)

6V1 + V2 = 3V1 + 3V2

3V1 = 2V2

or V1 = 0.67V2

If we want the resulting volume to be maximum, V2 should be maximum. The maximum available volume of V2 is 1 L.

Hence, V1 = 0.67 * 1 = 0.67 L

Therefore, the maximum volume of 3M KOH obtained will be V3 = V1 + V2 = 0.67 + 1 = 1.67L.

(2) By adding water

If we were allowed to add water, the maximum yield will be when we mix both the solutions (1M & 6M) and then add water, to dilute the solution to 3M. When we mix both, the resulting molarity will be:

MV = M1V1 + M2V2

M(2) = 6(1) + 1(1)

M = 3.5M

The solution will be 2L of 3.5M KOH. We have to dilute this, to bring it 3M

Again, applying the molarity equation, we have:

M1V1 + M2V2 = M3V3

If we add water, Molarity will be zero, hence M2 = 0

M1 = 3.5, V1 = 2L, M3 = 3 (since desired conc is 3M), V3 needs to be calculated

Therefore, 3.5 (2) = 3 (V3)

Hence, V3 = 7/3 = 2.33 L

Therefore, we get 2.33 L of 3M solution, by adding 0.33L of water to the mixture.

Hence, max volume of 3M KOH obtained is 2.33 L.

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