# Question #bca11

##### 1 Answer

#### Explanation:

This is a classic example of a dilution problem.

You know that **molarity**, *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

#V_1 = "250 cm"^3#

Your mixing this solution with

#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 = 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 *constant* is equivalent to **halving** the concentration of the solution.