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

## Question:

800 mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the solution to 1.9 M?

## Molarity:

In chemistry, the concentration of solutions can be expressed in more than one unit, such as molarity, normality and molality. However, molarity is commonly used. It describes the concentration in moles of solute per one liter of solution.

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.