Question #41fbf

1 Answer
Jul 5, 2015

Answer:

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"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

#520x = 300 * (x+ y)#

#520x - 300x = 300 y => y = (220 * x)/300 = 0.73 * 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.

For example, if you start with a 1-L solution, you need to add

#V_"add" = 0.73 * "1 L" = "0.73 L" = "730 mL"#

of water to go from 520 mOsmol/L to 300 mOsmol/L.

This of course implies that the volume of the final solution will be

#V_"final" = x + y = 1 + 0.73 = "1.73 L"#