A 50.0 mL volume of 0.50 M of X is mixed with 50.0 mL of 0.20 M of Y. a. Considering the value of K in the 100.0 mL of solution, determine the concentration of XY(aq)

0.1M

Explanation:

So, we are given the following data or parameters which is going to help in solving this particular Question:

=> There are two compounds/elements compound/element X and compound/element Y.

(NB: We are not giving the atomicity)

=> "A 50.0 mL volume of 0.50 M of X is mixed with 50.0 mL of 0.20 M of Y."

Thus, X + Y > XY(aq).

The first thing to do here is to determine the number of moles of each reactants that is X and Y.

The number of moles of X = 0.50M × 50 × 10^-3 L = 0.025 moles.

The number of moles of Y = 0.20M × 50 × 10^-3 L = 0.01 moles.

Hence, we have that 0.025 moles of X reacted with 0.01 moles of Y to give XY.

The limiting reagent/reactant is Y.

Therefore, 0.01 mole of Y × (1 mole XY/1 mole of Y) = 0.01 moles of XY.

Thus, the number of moles of XY is 0.01 mole.

Hence, we have that 0.025 moles of X reacted with 0.01 moles of Y to give 0.01 moles of XY.

So, the molarity or the concentration of XY(aq) = 0.01/(100 × 10^-3 ) = 0.1M

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