# Question 41fbf

Jul 5, 2015

That depends on what the volume of the initial solution is.

#### Explanation:

In order to be able to determine how much water is needed to reduce your solution's osmolarity from 520 mOsmol/L to 300 mOsmol/L, you need the actual volume of the initial solution.

Osmolarity expresses the number of osmoles per liters of solution.

In your case, you need to go from a more concentrated solution to a diluted solution. Let's assume that you have $x$ liters of your initial solution.

The number of osmoles this solution would contain would be equal to

$520 \text{mOsmol"/cancel("L") * xcancel("L") = (520 * x)" mOsmol}$

Now, let's assume that $y$ is the volume of water you need to add to this solution. The osmolarity of the second solution would be

((520 * x)"mOsmol")/((x + y)"L") = "300 mOsmol/L"#

This is equivalent to

$520 x = 300 \cdot \left(x + y\right)$

$520 x - 300 x = 300 y \implies y = \frac{220 \cdot x}{300} = 0.73 \cdot x$,

where $x$ is the volume of the initial solution.

So, to get how much water you need to add, simply multiply the initial volume by 0.73.

${V}_{\text{add" = 0.73 * "1 L" = "0.73 L" = "730 mL}}$
${V}_{\text{final" = x + y = 1 + 0.73 = "1.73 L}}$