800 mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the...

Given:

• Volume of the stock solution: {eq}\rm V_1 = 800~mL {/eq}

• Molarity of the stock solution: {eq}\rm c_1 = 3.0~M {/eq}

• Molarity of the diluted solution: {eq}\rm c_2 = 1.9~M {/eq}

Since the number of moles of the solute remains constant, and it can be expressed as a product of molarity {eq}\rm c {/eq} and volume {eq}\rm V {/eq}, we can state that:

$$\rm c_1V_1 = c_2V_2 $$

Rearranging for the volume of the diluted solution:

$$\rm V_2 = \dfrac{c_1V_1}{c_2} $$

Since this volume represents the sum of the volume of the stock solution {eq}\rm V_1 {/eq} and the volume of water added {eq}\rm V_{H_2O} {/eq}, we can state that:

$$\rm V_2 = V_1 + V_{H_2O} $$

Substituting this expression into the formula derived, we obtain:

$$\rm V_1 + V_{H_2O} = \dfrac{c_1V_1}{c_2} $$

Solving for the volume of water added:

$$\rm V_{H_2O} = \dfrac{c_1V_1}{c_2} - V_1\\ \implies V_{H_2O} = \dfrac{c_1V_1 - c_2V_1}{c_2}\\ \implies V_{H_2O} = \dfrac{V_1(c_1 - c_2)}{c_2} $$

Substituting the given data:

$$\rm V_{H_2O} = \dfrac{800~mL\cdot (3.0~M - 1.9~M)}{1.9~M}\\ \implies \boxed{\rm V_{H_2O} = 460~mL} $$

To summarize, 460 milliliters of water must be added to the original solution of 800 mL with the molarity of 1.9 M to dilute the solution to 1.9 M.