Question

A dilute solution of 0.05120 M was prepared when the lab temperature was 25 ^oC. On another day the solution was used when the temperature was 15 ^oC. What is the concentration of the solution at the 15 ^oC temperature?

Answer 1

We know that Molarity , M = number of moles of the solute / Volume of the solution in L

Given the molarity of the solution at 25 oC is 0.05120 M

Let the volume of the solution be 1L then molarity . M = number of moles / 1L = 0.05120 M

So number of moles of the solution is , n = 0.05120 M x 1L

                                                            = 0.05120 mol

We know that volume of the solution depend directly of the temperature of the solution,

So V is proportional to T

V / V' = T / T'

Where

V = initial volume at 25 oC = 1 L

V' = final volueme at 15oC = ?

T = initial temperature = 25 oC = 25+273 = 298 K

T' = final temperature = 15 oC = 15+273 = 288 K

Plug the values we get V' = (VxT') / T

                                     = (1L x 288K ) / 298K

                                    = 0.966 L

As the solute is the same the number of moles of the solute in the solution will not change so the New concentration of the solution is , M' = Number of moles of the solute / New volume

                                                   = 0.05120 mol / 0.966 L

                                                   = 0.05298 M

Therefore the concentration of the solution at the 15 ^oC temperature is 0.05298 M