Question #bca11

This is a classic example of a dilution problem.

You know that molarity, #c#, is defined as the number of moles of solute present per liter of solution. This can be expressed as

#c = n/V#

where

• #n# is the number of moles of solute
• #V# is the volume of the solution

Now, your initial solution has a concentration #c_1#, contains #n# moles of solute, and has a volume

#V_1 = "250 cm"^3#

Your mixing this solution with #"250 cm"^3# of water, which implies that the solution will still contain #n# moles of solute, but this time, the volume will be

#V_2 = "250 cm"^3 + "250 cm"^3 = "500 cm"^3 = color(red)(2) * V_1#

This means that the molarity of the solution, #c_2#, will be

#c_2 = n/V_2#

which is equal to

#c_2 = n/(color(red)(2) * V_1) = 1/color(red)(2) * n/V_1 = 1/color(red)(2) * c_1#

Therefore, you can say that

#c_2 = 1/color(red)(2) * "2 M" = "1 M"#

Long story short, doubling the volume of the solution while keeping the number of moles of solute, i.e. diluting the solution by a dilution factor of #2#, constant is equivalent to halving the concentration of the solution.