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.