Question #41fbf

1 Answer
Jul 5, 2015

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


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"#